Limit Et

34
7/28/2019 Limit Et http://slidepdf.com/reader/full/limit-et 1/34 Section 9: Presentation of the Theory toc 9. Presentation of the Theory In this section, we present the demonstration of many of the theorems discussed in the tutorial. Theorem 9.1. Let a and c be numbers, then lim x → a c = c. (Rule 1) Proof : Let > 0. In reference to the definition of limit, the function under consideration is f (x) = c, and L = c. We want to choose a number δ > 0 such that 0 < |x − a| < δ = ⇒ |f (x) − L| < or, 0 < |x − a| < δ = ⇒ |c − c| < (1) It is clear in this trivial situation that the condition | c − c | < will hold no matter the choice of δ > 0; therefore, choose δ = 1. Thus, for that choice of δ , obviously, (1) holds.

Transcript of Limit Et

Page 1: Limit Et

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Section 9 Presentation of the Theory toc9 Presentation of the Theory

In this section we present the demonstration of many of the theoremsdiscussed in the tutorial

Theorem 91 Let a and c be numbers then

limxrarra

c = c (Rule 1)

Proof Let gt 0 In reference to the definition of limit the functionunder consideration is f (x) = c and L = c We want to choose anumber δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt

or0 lt |x minus a| lt δ =rArr |c minus c| lt (1)

It is clear in this trivial situation that the condition|c

minusc|

lt willhold no matter the choice of δ gt 0 therefore choose δ = 1 Thus forthat choice of δ obviously (1) holds

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Section 9 Presentation of the Theory

Theorem 92 For any number a

limxrarra

x = a (Rule 2)

Proof Within the context of the definition of limit of a functionf (x) = x and L = a Let gt 0 We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt or

0 lt |x minus a| lt δ =rArr |x minus a| lt (2)

Towards that end choose δ = then

0 lt |x minus a| lt δ =rArr |x minus a| lt δ =

But this is exactly what we wanted to prove (2)

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Section 9 Presentation of the Theory

Theorem 93 (Algebra of Limits Theorem) Let f and g be functions

and let a and c be number Suppose

limxrarra

f (x) and limxrarra

g(x)

exist and are finite Then

(1) limxrarra

(f (x) + g(x)) = limxrarra

f (x) + limxrarra

g(x)

(2) limxrarra(cf (x)) = c limxrarra f (x)(3) lim

xrarra(f (x)g(x)) = lim

xrarraf (x) lim

xrarrag(x)

(4) limxrarra

f (x)

g(x)=

limxrarra

f (x)

limxrarra

g(x) provided lim

xrarrag(x) = 0

Throughout the proofs below let L = limxrarra

f (x) and M = limxrarra

g(x)

Proof of (1) Let gt 0 We must find a δ gt 0 such that

0 lt|x

minusa|

lt δ =rArr |

(f (x) + g(x))minus

(L + M )|

lt

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Section 9 Presentation of the Theory

Towards that end there exists a δ 2 gt 0 such that

0 lt|x

minusa|

lt δ 1 =rArr |

f (x)minus

L|

lt

2 (3)

since limxrarra

f (x) = L And there exists a δ 1 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2 (4)

Finally define δ = minδ 1 δ 2 Note that

0 lt |x minus a| lt δ =rArr 0 lt |x minus a| lt δ 1 and 0 lt |x minus a| lt δ 2 (5)

Now suppose x is a number in the domains of both f and g such that

0 lt |x minus a| lt δ (6)Then

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Section 9 Presentation of the Theory

|(f (x) + g(x)) minus (L + M )|= |(f (x) minus L) + (g(x) minus M )|le |f (x) minus L| + |g(x) minus M | Abs (1)

lt

2+

2(7)

=

Thus|(f (x) + g(x)) minus (L + M )| lt

Since we are assuming (6) the inequalities on the right-hand side of (5) Since those inequalities are obtained then the right-hand sides of (3) and (4) are true as well This is were we obtained the inequality

in (7)

We have shown that

0 lt |x minus a| lt δ =rArr |(f (x) + g(x)) minus (L + M )| lt

This is the definition of limxrarra(f (x) + g(x)) = L + M

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Section 9 Presentation of the Theory

Proof of (2) Let gt 0 We need to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |(cf (x))

minuscL

|lt (8)

Now since limxrarra

f (x) = L and gt 0 has been given from the definition

of limit there exists a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt

1 +|c

| (9)

Now suppose

0 lt |x minus a| lt δ

then

|(cf (x))

minuscL

|=

|c||

f (x)minus

L|

Abs (3)

lt |c|

1 + |c|=

|c|1 +

|c

|

le Thus

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Section 9 Presentation of the Theory

|(cf (x)) minus cL| lt

Note We utilized the fact that |c

|1 + |c| le 1 for any number c isin RThe denominator 1 + |c| was chosen instead of |c| to account for thepossibility that c might be zero

Thus we have shown that for the δ gt 0 produced in (9) (8) is ob-

tained

Proof of (3) Let gt 0 be given We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt (10)

Consider the following series of manipulations

|f (x)g(x) minus LM |= |f (x)g(x) minus f (x)M + f (x)M minus LM |

le |f (x)g(x) minus f (x)M | + |f (x)M minus LM |

Abs (1)= |f (x)||g(x) minus M | + |M ||f (x) minus L| Abs (3) (11)

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 2: Limit Et

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Section 9 Presentation of the Theory

Theorem 92 For any number a

limxrarra

x = a (Rule 2)

Proof Within the context of the definition of limit of a functionf (x) = x and L = a Let gt 0 We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt or

0 lt |x minus a| lt δ =rArr |x minus a| lt (2)

Towards that end choose δ = then

0 lt |x minus a| lt δ =rArr |x minus a| lt δ =

But this is exactly what we wanted to prove (2)

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Section 9 Presentation of the Theory

Theorem 93 (Algebra of Limits Theorem) Let f and g be functions

and let a and c be number Suppose

limxrarra

f (x) and limxrarra

g(x)

exist and are finite Then

(1) limxrarra

(f (x) + g(x)) = limxrarra

f (x) + limxrarra

g(x)

(2) limxrarra(cf (x)) = c limxrarra f (x)(3) lim

xrarra(f (x)g(x)) = lim

xrarraf (x) lim

xrarrag(x)

(4) limxrarra

f (x)

g(x)=

limxrarra

f (x)

limxrarra

g(x) provided lim

xrarrag(x) = 0

Throughout the proofs below let L = limxrarra

f (x) and M = limxrarra

g(x)

Proof of (1) Let gt 0 We must find a δ gt 0 such that

0 lt|x

minusa|

lt δ =rArr |

(f (x) + g(x))minus

(L + M )|

lt

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Section 9 Presentation of the Theory

Towards that end there exists a δ 2 gt 0 such that

0 lt|x

minusa|

lt δ 1 =rArr |

f (x)minus

L|

lt

2 (3)

since limxrarra

f (x) = L And there exists a δ 1 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2 (4)

Finally define δ = minδ 1 δ 2 Note that

0 lt |x minus a| lt δ =rArr 0 lt |x minus a| lt δ 1 and 0 lt |x minus a| lt δ 2 (5)

Now suppose x is a number in the domains of both f and g such that

0 lt |x minus a| lt δ (6)Then

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Section 9 Presentation of the Theory

|(f (x) + g(x)) minus (L + M )|= |(f (x) minus L) + (g(x) minus M )|le |f (x) minus L| + |g(x) minus M | Abs (1)

lt

2+

2(7)

=

Thus|(f (x) + g(x)) minus (L + M )| lt

Since we are assuming (6) the inequalities on the right-hand side of (5) Since those inequalities are obtained then the right-hand sides of (3) and (4) are true as well This is were we obtained the inequality

in (7)

We have shown that

0 lt |x minus a| lt δ =rArr |(f (x) + g(x)) minus (L + M )| lt

This is the definition of limxrarra(f (x) + g(x)) = L + M

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Section 9 Presentation of the Theory

Proof of (2) Let gt 0 We need to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |(cf (x))

minuscL

|lt (8)

Now since limxrarra

f (x) = L and gt 0 has been given from the definition

of limit there exists a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt

1 +|c

| (9)

Now suppose

0 lt |x minus a| lt δ

then

|(cf (x))

minuscL

|=

|c||

f (x)minus

L|

Abs (3)

lt |c|

1 + |c|=

|c|1 +

|c

|

le Thus

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Section 9 Presentation of the Theory

|(cf (x)) minus cL| lt

Note We utilized the fact that |c

|1 + |c| le 1 for any number c isin RThe denominator 1 + |c| was chosen instead of |c| to account for thepossibility that c might be zero

Thus we have shown that for the δ gt 0 produced in (9) (8) is ob-

tained

Proof of (3) Let gt 0 be given We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt (10)

Consider the following series of manipulations

|f (x)g(x) minus LM |= |f (x)g(x) minus f (x)M + f (x)M minus LM |

le |f (x)g(x) minus f (x)M | + |f (x)M minus LM |

Abs (1)= |f (x)||g(x) minus M | + |M ||f (x) minus L| Abs (3) (11)

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

Theorem 93 (Algebra of Limits Theorem) Let f and g be functions

and let a and c be number Suppose

limxrarra

f (x) and limxrarra

g(x)

exist and are finite Then

(1) limxrarra

(f (x) + g(x)) = limxrarra

f (x) + limxrarra

g(x)

(2) limxrarra(cf (x)) = c limxrarra f (x)(3) lim

xrarra(f (x)g(x)) = lim

xrarraf (x) lim

xrarrag(x)

(4) limxrarra

f (x)

g(x)=

limxrarra

f (x)

limxrarra

g(x) provided lim

xrarrag(x) = 0

Throughout the proofs below let L = limxrarra

f (x) and M = limxrarra

g(x)

Proof of (1) Let gt 0 We must find a δ gt 0 such that

0 lt|x

minusa|

lt δ =rArr |

(f (x) + g(x))minus

(L + M )|

lt

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Section 9 Presentation of the Theory

Towards that end there exists a δ 2 gt 0 such that

0 lt|x

minusa|

lt δ 1 =rArr |

f (x)minus

L|

lt

2 (3)

since limxrarra

f (x) = L And there exists a δ 1 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2 (4)

Finally define δ = minδ 1 δ 2 Note that

0 lt |x minus a| lt δ =rArr 0 lt |x minus a| lt δ 1 and 0 lt |x minus a| lt δ 2 (5)

Now suppose x is a number in the domains of both f and g such that

0 lt |x minus a| lt δ (6)Then

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Section 9 Presentation of the Theory

|(f (x) + g(x)) minus (L + M )|= |(f (x) minus L) + (g(x) minus M )|le |f (x) minus L| + |g(x) minus M | Abs (1)

lt

2+

2(7)

=

Thus|(f (x) + g(x)) minus (L + M )| lt

Since we are assuming (6) the inequalities on the right-hand side of (5) Since those inequalities are obtained then the right-hand sides of (3) and (4) are true as well This is were we obtained the inequality

in (7)

We have shown that

0 lt |x minus a| lt δ =rArr |(f (x) + g(x)) minus (L + M )| lt

This is the definition of limxrarra(f (x) + g(x)) = L + M

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Section 9 Presentation of the Theory

Proof of (2) Let gt 0 We need to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |(cf (x))

minuscL

|lt (8)

Now since limxrarra

f (x) = L and gt 0 has been given from the definition

of limit there exists a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt

1 +|c

| (9)

Now suppose

0 lt |x minus a| lt δ

then

|(cf (x))

minuscL

|=

|c||

f (x)minus

L|

Abs (3)

lt |c|

1 + |c|=

|c|1 +

|c

|

le Thus

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Section 9 Presentation of the Theory

|(cf (x)) minus cL| lt

Note We utilized the fact that |c

|1 + |c| le 1 for any number c isin RThe denominator 1 + |c| was chosen instead of |c| to account for thepossibility that c might be zero

Thus we have shown that for the δ gt 0 produced in (9) (8) is ob-

tained

Proof of (3) Let gt 0 be given We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt (10)

Consider the following series of manipulations

|f (x)g(x) minus LM |= |f (x)g(x) minus f (x)M + f (x)M minus LM |

le |f (x)g(x) minus f (x)M | + |f (x)M minus LM |

Abs (1)= |f (x)||g(x) minus M | + |M ||f (x) minus L| Abs (3) (11)

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

Towards that end there exists a δ 2 gt 0 such that

0 lt|x

minusa|

lt δ 1 =rArr |

f (x)minus

L|

lt

2 (3)

since limxrarra

f (x) = L And there exists a δ 1 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2 (4)

Finally define δ = minδ 1 δ 2 Note that

0 lt |x minus a| lt δ =rArr 0 lt |x minus a| lt δ 1 and 0 lt |x minus a| lt δ 2 (5)

Now suppose x is a number in the domains of both f and g such that

0 lt |x minus a| lt δ (6)Then

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Section 9 Presentation of the Theory

|(f (x) + g(x)) minus (L + M )|= |(f (x) minus L) + (g(x) minus M )|le |f (x) minus L| + |g(x) minus M | Abs (1)

lt

2+

2(7)

=

Thus|(f (x) + g(x)) minus (L + M )| lt

Since we are assuming (6) the inequalities on the right-hand side of (5) Since those inequalities are obtained then the right-hand sides of (3) and (4) are true as well This is were we obtained the inequality

in (7)

We have shown that

0 lt |x minus a| lt δ =rArr |(f (x) + g(x)) minus (L + M )| lt

This is the definition of limxrarra(f (x) + g(x)) = L + M

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Section 9 Presentation of the Theory

Proof of (2) Let gt 0 We need to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |(cf (x))

minuscL

|lt (8)

Now since limxrarra

f (x) = L and gt 0 has been given from the definition

of limit there exists a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt

1 +|c

| (9)

Now suppose

0 lt |x minus a| lt δ

then

|(cf (x))

minuscL

|=

|c||

f (x)minus

L|

Abs (3)

lt |c|

1 + |c|=

|c|1 +

|c

|

le Thus

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Section 9 Presentation of the Theory

|(cf (x)) minus cL| lt

Note We utilized the fact that |c

|1 + |c| le 1 for any number c isin RThe denominator 1 + |c| was chosen instead of |c| to account for thepossibility that c might be zero

Thus we have shown that for the δ gt 0 produced in (9) (8) is ob-

tained

Proof of (3) Let gt 0 be given We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt (10)

Consider the following series of manipulations

|f (x)g(x) minus LM |= |f (x)g(x) minus f (x)M + f (x)M minus LM |

le |f (x)g(x) minus f (x)M | + |f (x)M minus LM |

Abs (1)= |f (x)||g(x) minus M | + |M ||f (x) minus L| Abs (3) (11)

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

7282019 Limit Et

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 5: Limit Et

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Section 9 Presentation of the Theory

|(f (x) + g(x)) minus (L + M )|= |(f (x) minus L) + (g(x) minus M )|le |f (x) minus L| + |g(x) minus M | Abs (1)

lt

2+

2(7)

=

Thus|(f (x) + g(x)) minus (L + M )| lt

Since we are assuming (6) the inequalities on the right-hand side of (5) Since those inequalities are obtained then the right-hand sides of (3) and (4) are true as well This is were we obtained the inequality

in (7)

We have shown that

0 lt |x minus a| lt δ =rArr |(f (x) + g(x)) minus (L + M )| lt

This is the definition of limxrarra(f (x) + g(x)) = L + M

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Section 9 Presentation of the Theory

Proof of (2) Let gt 0 We need to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |(cf (x))

minuscL

|lt (8)

Now since limxrarra

f (x) = L and gt 0 has been given from the definition

of limit there exists a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt

1 +|c

| (9)

Now suppose

0 lt |x minus a| lt δ

then

|(cf (x))

minuscL

|=

|c||

f (x)minus

L|

Abs (3)

lt |c|

1 + |c|=

|c|1 +

|c

|

le Thus

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Section 9 Presentation of the Theory

|(cf (x)) minus cL| lt

Note We utilized the fact that |c

|1 + |c| le 1 for any number c isin RThe denominator 1 + |c| was chosen instead of |c| to account for thepossibility that c might be zero

Thus we have shown that for the δ gt 0 produced in (9) (8) is ob-

tained

Proof of (3) Let gt 0 be given We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt (10)

Consider the following series of manipulations

|f (x)g(x) minus LM |= |f (x)g(x) minus f (x)M + f (x)M minus LM |

le |f (x)g(x) minus f (x)M | + |f (x)M minus LM |

Abs (1)= |f (x)||g(x) minus M | + |M ||f (x) minus L| Abs (3) (11)

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 6: Limit Et

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Section 9 Presentation of the Theory

Proof of (2) Let gt 0 We need to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |(cf (x))

minuscL

|lt (8)

Now since limxrarra

f (x) = L and gt 0 has been given from the definition

of limit there exists a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x) minus L| lt

1 +|c

| (9)

Now suppose

0 lt |x minus a| lt δ

then

|(cf (x))

minuscL

|=

|c||

f (x)minus

L|

Abs (3)

lt |c|

1 + |c|=

|c|1 +

|c

|

le Thus

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Section 9 Presentation of the Theory

|(cf (x)) minus cL| lt

Note We utilized the fact that |c

|1 + |c| le 1 for any number c isin RThe denominator 1 + |c| was chosen instead of |c| to account for thepossibility that c might be zero

Thus we have shown that for the δ gt 0 produced in (9) (8) is ob-

tained

Proof of (3) Let gt 0 be given We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt (10)

Consider the following series of manipulations

|f (x)g(x) minus LM |= |f (x)g(x) minus f (x)M + f (x)M minus LM |

le |f (x)g(x) minus f (x)M | + |f (x)M minus LM |

Abs (1)= |f (x)||g(x) minus M | + |M ||f (x) minus L| Abs (3) (11)

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

7282019 Limit Et

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 7: Limit Et

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Section 9 Presentation of the Theory

|(cf (x)) minus cL| lt

Note We utilized the fact that |c

|1 + |c| le 1 for any number c isin RThe denominator 1 + |c| was chosen instead of |c| to account for thepossibility that c might be zero

Thus we have shown that for the δ gt 0 produced in (9) (8) is ob-

tained

Proof of (3) Let gt 0 be given We want to find a δ gt 0 such that

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt (10)

Consider the following series of manipulations

|f (x)g(x) minus LM |= |f (x)g(x) minus f (x)M + f (x)M minus LM |

le |f (x)g(x) minus f (x)M | + |f (x)M minus LM |

Abs (1)= |f (x)||g(x) minus M | + |M ||f (x) minus L| Abs (3) (11)

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

We can make the second term in (11) ldquosmallrdquo since limxrarra

f (x) = L

Choose δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |f (x) minus L| lt

2(1 + |M | (12)

Now for the problem of the first term of (11) We can make the factor

|g(x)

minusM

|as small as we wish since limxrarra g(x) = M but we have

to make sure that the smallest of this factor is not counter-balancedby the factor |f (x)| (That factor might be large mdash we have to makesure that it is not) To that end for 0 lt |x minus a| lt δ 1 we have by(12)

|f (x) minus L| lt 2(1 + |M |)

then|f (x)| = |f (x) minus L + L| le |f (x) minus L| + |L| Abs (1)

le

2(1 + |M | + |L|)

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

7282019 Limit Et

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

For the purpose of convenience let C =

2(1 + |M | + |L|) Thus thereis a constant C such that

0 lt |x minus a| lt δ 1 =rArr |f (x)| lt C (13)

Continuing our quest for the ultimate δ gt 0 for which (10) is validchoose δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |g(x) minus M | lt

2(1 + C ) (14)

This is possible since limxrarra

g(x) = M

Finally choose the ultimate δ -value as

δ = minδ 1 δ 2 (15)

and suppose 0 lt |xminusa| lt δ Then 0 lt |xminusa| lt δ 1 and 0 lt |xminusa| lt δ 2form (15) and as a consequence the inequalities in (12) (14) and (13)

S i 9 P i f h Th

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

7282019 Limit Et

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

are valid Thus for 0 lt |x minus a| lt δ we have

|f (x)g(x)

minusLM

|le |f (x)||g(x) minus M | + |M ||f (x) minus L| (11)

lt C

2(1 + C )+ |M |

2(1 + |M |) (13) (14) (12)

lt

C

1 + C

2 + |M

|1 + |M |

2

le

Thus we have shown that for any gt 0 there exists a δ gt 0 suchthat

0 lt |x minus a| lt δ =rArr |f (x)g(x) minus LM | lt

This completes the proof of this part

Proof of (4) Let gt 0 From the definition we want to find a numberδ gt 0 such that

0 lt |x minus a| lt δ =rArrf (x)

g(x)minus L

M

lt

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

7282019 Limit Et

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

7282019 Limit Et

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

7282019 Limit Et

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

here we are assuming M = 0

Towards this end consider the followingf (x)

g(x)minus L

M

=

f (x)M minus g(x)L

M g(x)

=

(f (x)M minus LM ) + (LM minus g(x)L)

M g(x)

=

M (f (x) minus L) + L(M minus g(x))

M g(x)

le

M (f (x) minus L)

M g(x)

+

L(M minus g(x))

M g(x)

Abs (1)

=1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | Abs (3)

Do you see where I am going with this calculation To summarize wehave shown thatf (x)

g(x)minus L

M

le 1

|g(x)| |f (x) minus L| +|L|

|M g(x)| |g(x) minus M | (16)

S ti 9 P t ti f th Th

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

Now for some details The |g(x)| in the denominator is bothersomeSince lim

xrarrag(x) = M = 0 there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus M | ltM

2 definition

Now for 0 lt |x minus a| lt δ 1

|g(x)

|=

|M

minus(M

minusg(x))

|ge |M | minus |M minus g(x)| Abs (2)

ge |M | minus M

2

geM

2

This establishes the inequality

0 lt |x minus a| lt δ 1 =rArr |g(x)| ge M

2

=rArr 1|g(x)| le 2M

(18)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

7282019 Limit Et

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

Letrsquos put this back into the inequality (16) to get

f (x)

g(x) minusL

M le1

|g(x)| |f (x) minus L| + |L

||M g(x)| |g(x) minus M |le 2

|M | |f (x) minus L| +2|L|M 2

|g(x) minus M | (20)

Now for our final calculations Since limxrarra f (x) = L there is a δ 2 gt 0such that

0 lt |x minus a| lt δ 2 =rArr |f (x) minus L| lt|M |

4 (21)

Since limxrarrag

(x

) =M

there is aδ 3

gt0 such that

0 lt |x minus a| lt δ 3 =rArr |g(x) minus M | ltM 2

4(1 + |L|) (22)

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

7282019 Limit Et

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

Choose δ = minδ 1 δ 2 δ 3 Consequently if x satisfies the inequality0 lt |x minus a| lt δ then all three of the inequalities (18) (21) and (22)

hold Thus if we take x to satisfy0 lt |x minus a| lt δ

then from (20) we have

f (x)

g(x) minusL

M le2

|M | |f (x) minus L| +

2

|L

|M 2 |g(x) minus M |le 2

|M ||M |

4+

2|L|M 2

M 2

4(1 + |L|) from (21) and (22)

le

2+

2=

That does it We have found the δ gt 0 that ldquoworksrdquo

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

Theorem 94 (Continuity of Power Functions) Let a isin R and n isinN Then

limxrarra x

n

= a

n

(23)

Proof By Theorem 92 we have limxrarra

x = a Therefore by Theo-

rem 93 (3)

limxrarra x2 = limxrarra xx = limxrarra x limxrarra x = aa = a2

Similarly by Theorem 93 (3)

limxrarra

x2 = limxrarra

x2x = limxrarra

x2 limxrarra

x = a2a = a3

The formal mechanism for finishing the proof in Mathemtical Induc-tion Suppose we have shown that

limxrarra

xnminus1 = anminus1

then from Theorem 93 (3) and Theorem 92limxrarra

xn = limxrarra

xnminus1x = limxrarra

xnminus1 limxrarra

x = anminus1a = an

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

By the Principle of Mathematical Induction we have proved the the-orem

Theorem 95 (Continuity of Polynomial Functions) Let p be a poly-

nomial and a isin R Then

limxrarra

p(x) = p(a) (24)

Proof Let p(x) be a polynomial of degree n isin N This means thatthe functional form of p is

p(x) = b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn

for some set of coefficients b0 b1 b2 bn

Notice that p(a)b0 + b1a + b2a2 + b3a3 + middot middot middot + bnminus1anminus1 + bnan

Section 9 Presentation of the Theory

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Section 9 Presentation of the Theory

Now apply Theorem 93 (1) Theorem 93 (2) as well as Theorem 94to obtain

limxrarra

p(x) = limxrarra(b0 + b1x + b2x2 + b3x3 + middot middot middot + bnminus1xnminus1 + bnxn)

= limxrarra

b0 + limxrarra

b1x + limxrarra

b2x2 + middot middot middot + limxrarra

bnxn (1)

= b0 + b1 limxrarra

x + b2 limxrarra

x2 + middot middot middot + bn limxrarra

xn (2)Rule 1

= b0 + b1a + b2a2 + middot middot middot + bnan Thm 94

= p(a)

Theorem 96 (Continuity of Rational Functions) Let f be a rational

function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a) (25)

Section 9 Presentation of the Theory

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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S 9 s y

Proof A rational function is a quotient of two polynomials Let p andq be two polynomials such that

f (x) = p(x)q (x)

The natural domain of f is given by

Dom(f ) =

x

isinR

|g(x)

= 0

Let a isin Dom(f ) thus g(a) = 0 This observation is important as weare about to cite Theorem 93 (4) In that theorem we require thelimit of the denominator to be nonzero Read on

limxrarra f (x) = limxrarra

p(x)

q (x)

=limxrarra

p(x)

limxrarra

q (x) Thm 93 (4)

= p(a)q (a)

Thm 95

Section 9 Presentation of the Theory

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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y

Thus

limxrarra

f (x) =p(a)

q (a)

= f (a)

This is equation (25) that was asserted in the theorem

Theorem 97 Let f and g be functions that are compatible for com-

position let a

isinR Suppose

(1) limxrarra g(x) exists let b = limxrarra g(x)(2) b isin Dom(f ) and limyrarrb f (y) = f (b) exists

Then

limxrarra

f (g(x)) exists

and

limxrarra

f (g(x)) = f (b)

or

limxrarra

f (g(x)) = f (limxrarra

g(x)) (26)

Section 9 Presentation of the Theory

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

httpslidepdfcomreaderfulllimit-et 3234

Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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y

Proof Let gt 0 We want to find a number δ gt 0 such that

0 lt

|x

minusa

|lt δ =

rArr |f (g(x))

minusf (b)

|lt (27)

To this end since gt 0 has been given and it is assumed thatlimyrarrb

f (y) = f (b) there is a γ gt 0 such that

0 lt |y minus b| lt γ =rArr |f (y) minus f (b)| lt (28)

Now we have a number γ gt 0 defined since we are assuming thatlimxrarra

g(x) = b there is a number δ gt 0 such that

0 lt |x minus a| lt δ =rArr |g(x) minus b| lt γ (29)

Now we claim that the number δ gt 0 produced in the last paragraphis the δ -value we seek Indeed suppose

0 lt |x minus a| lt δ

Then from (29)

|g(x) minus b| lt γ

Section 9 Presentation of the Theory

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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But now this inequality implies by (28) that

|f (g(x))

minusf (b)

|lt

This proves the theorem

Theorem 98 (Continuity of the Root Function) Let n isin N Define

f (x) = n

radic x for a isin Dom(f ) Then

limxrarra

f (x) = f (a)or

limxrarra

n

radic x = n

radic a (30)

Proof The key to this proof is the observation that f (x) =nradic x is an

increasing function ie

x1 x2 isin Dom(f ) and x1 lt x2 =rArr f (x1) lt f (x2)or

x1 x2 isin Dom(f ) and x1 lt x2 =rArrn

radic x1 ltn

radic x2

Section 9 Presentation of the Theory

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

7282019 Limit Et

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Let a isin Dom(f ) Show limxrarran

radic x = n

radic a

Case I a = 0 Let gt 0 choose δ = n Then it is obvious that sincen

radic 0 = 0

x isin Dom(f ) and |x| lt δ = n =rArr | nradic x| lt

This is the trivial case

Case II a gt 0 Let gt 0 We can assume is so small thatn

radic aminus gt 0(Why) Then

0 lt n

radic a minus lt n

radic a lt n

radic a + =rArr 0 lt ( n

radic a minus )n lt a lt ( n

radic a + )n

This is because the function x rarr xn is increasing on the interval

( 0 infin ) ie 0 lt x1 lt x2 =rArr xn1 lt x

n2

Define the δ gt 0 we are looking for as

δ = min a minus ( nradic

a minus )n ( nradic

a + )n minus a

Section 9 Presentation of the Theory

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

7282019 Limit Et

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Now suppose 0 lt |x minus a| lt δ Notice that such an x must necessarilybelong to the domain of f (Why) Then

0 lt |x minus a| lt δ =rArr a minus δ lt x lt a + δ

=rArr ( nradic

a minus )n lt a minus δ lt x lt a + δ lt ( nradic

a + )n

(32)

Thus we have

0 lt |x minus a| lt δ =rArr ( nradic a minus )n lt x lt ( nradic a + )n

Now we use the increasing property of nradic

n

0 lt |x minus a| lt δ =rArr ( nradic

a minus )n lt x lt ( nradic

a + )n

=rArr n ( nradic a minus n) ltnradic x lt

n ( nradic a + n)

=rArr n

radic a minus lt n

radic x lt n

radic a +

=rArr | nradic x minus n

radic a| lt

Thus we have shown that0 lt |x minus a| lt δ =rArr | nradic

x minus n

radic a| lt (33)

Section 9 Presentation of the Theory

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

7282019 Limit Et

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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This proves that limxrarran

radic x = n

radic a in this case

Case III a lt 0 This case is only present with n the root being

extracted is an odd integer The proof is left to the reader mdash it issimilar to Case II The student need only study the previous casemake appropriate changes in the steps

Proof Notes How can we assume that is such that n

radic a minus gt 0 If

does not satisfy this inequality then choose another -value say 1that does satisfy the desired inequality Now use 1 throughout therest of the proof instead of We would finished the proof with thedeclaration that

0 lt

|x

minusa

|lt δ =

rArr |n

radic x

minusn

radic a

|lt 1

but since 1 lt we would have had

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt 1 lt

which is (33) That being the case we might as well assume at the

beginning that a minus gt 0 and avoid the introduction of 1

Section 9 Presentation of the Theory

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

7282019 Limit Et

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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If you go to the trouble of studying the proof given in Case II it would appear that the only property of n

radic x used in the proof was

that it was increasing (I used that fact that xrarr

xn is increasing aswell) That being the case can this proof be modified to argue thatany increasing function f has the property that limxrarra f (x) = f (a)

Having read the previous paragraph now consider the function

f (x) = x x lt 0

1 + x x ge 0

This function is strictly increasing over R but limxrarr0 f (x) = f (0)since the two-sided limit does not exist

Can you resolve the seeming contradiction between the two

paragraphs e-mail me with your thoughtsExercise 91 Draw the graph of f (x) = n

radic x and use it to illustrate

the main idea of the proof of Case II of Theorem 98

Exercise 92 Let n

isinN be odd and let a

isinR be negative Prove

limxrarra

nradic x = nradic a

Section 9 Presentation of the Theory

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

7282019 Limit Et

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

7282019 Limit Et

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

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Corollary 99 Suppose limxrarra

g(x) exists then

limxrarra

n g(x) =n

radic limxrarra

g(x) (34)

provided that the number b = limxrarra

g(x) is within the domain of the

nth-root function

Proof All the heavy lifting has been done We apply Theorem 97with the function f in that theorem as f (x) = nradic x and the functiong in that theorem the function g in this corollary Now the point of Theorem 98 was that it is a verification of condition (2) Of coursecondition (1) is apart of the assumptions of this corollary Therefore

we can conclude by Theorem 97 thatlimxrarra

n

g(x) =

n

radic limxrarra

g(x)

which is (34)

Section 9 Presentation of the Theory

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 27: Limit Et

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Theorem 910 (Continuity of Algebraic Functions) Let f be an al-

gebraic function and let a isin Dom(f ) Then

limxrarra

f (x) = f (a)

Proof An algebraic function is constructed by sums differences prod-ucts quotients and compositions with functions of the form

y = c y = xn y = mradic x

These are constant functions power functions and root functionsrespectively For all three we have shown the property

limxrarra

c = c limxrarra

xn = an limxrarra

m

radic x = m

radic a

The latter case is true provided a belongs to the domain of the mth-root functions (See theorems 91 94 and 98)

These observations combined with the Algebra of Limits Theorem

which concerns sums differences products and quotients of functions

Section 9 Presentation of the Theory

7282019 Limit Et

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

7282019 Limit Et

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 28: Limit Et

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whose limits exist and with Theorem 97 which concerns composi-tions of functions whose limits exist allows us to make the assertionof the theorem

Theorem 911 Let g f and h be functions and a L isin R Suppose

there is some δ 0 gt 0 such that

g(x)le

f (x)le

h(x)|x

minusa

|lt δ 0 (35)

and

limxrarra

g(x) = limxrarra

h(x) = L

Then

limxrarr

a

f (x) = L

Proof Let gt 0

Since limxrarra g(x) = L there is a δ 1 gt 0 such that

0 lt |x minus a| lt δ 1 =rArr |g(x) minus L| lt

Section 9 Presentation of the Theory

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

7282019 Limit Et

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 29: Limit Et

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This inequality implies

L

minus lt g(x) whenever 0 lt

|x

minusa

|lt δ 1 (36)

(Do you know why)

Now since limxrarra h(x) = L there is a δ 2 gt 0 such that

0 lt |x minus a| lt δ 2 =rArr |h(x) minus L| lt

This implies

h(x) lt L + whenever 0 lt |x minus a| lt δ 2 (37)

(Do you know why)

Finally choose δ = min δ 0 δ 1 δ 2 then

0 lt |x minus a| lt δ

impliesL minus lt

(36)

g(x) le f (x) le h(x)

(35)

lt L +

(37)

Section 9 Presentation of the Theory

7282019 Limit Et

httpslidepdfcomreaderfulllimit-et 3034

But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

7282019 Limit Et

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

httpslidepdfcomreaderfulllimit-et 3234

Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

7282019 Limit Et

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 30: Limit Et

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But this implies

|f (x) minus L| lt whenever 0 lt |x minus a| lt δ

which is what we wanted to prove

Theorem 912 The following limits are obtained

limxrarr

0

sin(x) = 0 limxrarr

0

cos(x) = 1 (38)

Proof Under construction

Theorem 913 The following limits are obtained

limxrarr0

sin(x)x

= 1 limxrarr0

1 minus cos(x)x

= 0 (39)

Proof Under construction

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

httpslidepdfcomreaderfulllimit-et 3234

Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

7282019 Limit Et

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 31: Limit Et

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Appendix

Properties of Absolute Value Let a b c isin R then each of thefollowing inequalities are obtained

1 |a + b| le |a| + |b|2 |a minus b| ge |a| minus |b|3

|ab

|=

|a

||b

|

7282019 Limit Et

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

7282019 Limit Et

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 32: Limit Et

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Solutions to Exercises

91 Draw a graph that represents f Designate a point on the hori-zontal axis as a then mark off the corresponding point on the verticalaxis mdash label this point n

radic a

Now mark off a little distance equidistant above and below the

point labeledn

radic a These are the pointsn

radic a minus andn

radic a + Labelthem so

Now starting at each of the points n

radic a minus and n

radic a + move hori-

zontally until you hit the graph of f now move vertically downwarduntil you intersect the x-axis The two points obtained in this wayare ( nradic a minus )n and ( nradic a + )n respectively These two points have thenumber a between them

The quantity δ is the shortest distance between a and each of thetwo points constructed in the previous paragraph Now mark off a

δ -distance on either side of a Do you see that this δ interval lies com-pletely inside the larger interval (Because δ is the shortest distance

Solutions to Exercises (continued)

7282019 Limit Et

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 33: Limit Et

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to the endpoints) Now it is clear at least geometrically the validityof the inequality (32) Can you show the inequality (32) algebraically

Assuming you understand my verbal (bit-tell) instructions do yousee why

0 lt |x minus a| lt δ =rArr | nradic x minus n

radic a| lt

Exercise Notes Review my question in the Proof Notes above Does

this graphical construction give you additional insight into answeringthe question

Exercise 91

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92

Page 34: Limit Et

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92 I said mdash it is left to the reader DPS Study the proof of Case II of Theorem 98 and make appropriate

changes to correspond to a lt 0Exercise 92


TEST METHOD FOR LIQUID LIMIT, PLASTIC LIMIT, AND ...

TEST METHOD FOR LIQUID LIMIT, PLASTIC LIMIT, AND ...

Karar Limiti–Tespit limiti (Decision limit = Limit of ...

Karar Limiti–Tespit limiti (Decision limit = Limit of ...

Standard Penetration Test & Liquid Limit,Plasticity Limit

Standard Penetration Test & Liquid Limit,Plasticity Limit

Limit of detection, limit of quantification and limit of blank Elvar Theodorsson.

Limit of detection, limit of quantification and limit of blank Elvar Theodorsson.

Physiologie et Monitorage - anesthesiologie.umontreal.ca · Cours de sciences de base ... • Mécanisme secondaire et limit ... – Déplacement du contenu intracrânien

Physiologie et Monitorage - anesthesiologie.umontreal.ca · Cours de sciences de base ... • Mécanisme secondaire et limit ... – Déplacement du contenu intracrânien

An upper limit for slow-earthquake zones: self-oscillatory ... · ties of Dieterich–Ruina friction law (Dieterich ,1979;Ruina 1983;Watkins et al.,2015;Marone et al.,2015;Scuderi

An upper limit for slow-earthquake zones: self-oscillatory ... · ties of Dieterich–Ruina friction law (Dieterich ,1979;Ruina 1983;Watkins et al.,2015;Marone et al.,2015;Scuderi

Spray | 2oz Limit 3 Limit 2 Limit 3 spread

Spray | 2oz Limit 3 Limit 2 Limit 3 spread

Limit 99 · 2020-07-04 · Limit 4 BUY 1 GET FREE 1 EQUAL OR LESSER VALUE Limit 3 Limit 12 Limit 4 Limit 4 Limit 4 Limit 2 Some ad items may vary by store and are available while

Limit 99 · 2020-07-04 · Limit 4 BUY 1 GET FREE 1 EQUAL OR LESSER VALUE Limit 3 Limit 12 Limit 4 Limit 4 Limit 4 Limit 2 Some ad items may vary by store and are available while

Central limit theorems for additive functionals of ergodic ... · morowski et al. (2012) and Ho¨pfner and Lo¨cherbach (2003) for null recurrent Markov processes. Central limit theorems

Central limit theorems for additive functionals of ergodic ... · morowski et al. (2012) and Ho¨pfner and Lo¨cherbach (2003) for null recurrent Markov processes. Central limit theorems

KALKULUS DASAR - · PDF file• Teorema Limit • Limit Kiri dan :Limit Kanan • Limit Tak Hingga • Kekontinuan Fungsi IV. TURUNAN FUNGSI ... 2.3 Pengertian Limit

KALKULUS DASAR - · PDF file• Teorema Limit • Limit Kiri dan :Limit Kanan • Limit Tak Hingga • Kekontinuan Fungsi IV. TURUNAN FUNGSI ... 2.3 Pengertian Limit

Vespula vulgaris) poisoning using clustered bait stations ... · not limit peak wasp abundance or ecological damage (Moller et al. 1991a; Beggs et al. 1998). Moreover, the level of

Vespula vulgaris) poisoning using clustered bait stations ... · not limit peak wasp abundance or ecological damage (Moller et al. 1991a; Beggs et al. 1998). Moreover, the level of

Le No Limit Heads Up Hold’em et l’apprentissage de la machine

Le No Limit Heads Up Hold’em et l’apprentissage de la machine

Radio Loud andRadio Quiet QSOs - arXiv · with a 1mJybeam−1 catalog limit, FIRST cannot detect RQQs individually, so White et al. (2007) lowered their noise limit by stacking over

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