MTAEA Continuity and Limits of Functions · Continuity and Limits of Functions Continuous...
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Continuity and Limits of Functions
MTAEA – Continuity and Limits of Functions
Scott McCracken
School of Economics,Australian National University
February 1, 2010
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Continuous Functions.
I A continuous function is one we can draw without taking our penoff the paper
Definition.Let f be a real-valued function whose domain is a subset of R.
I The function f is continuous at x0 in dom(f ) if, for every sequence(xn) in dom(f ) converging to x0, we have lim f (xn) = f (x0).
I If f is continuous at each point of a set S ⊆ dom(f ), then f is saidto be continuous on S.
I The function f is said to be continuous if it is continuous ondom(f ). N
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Continuous Functions.
x
y
x0
Figure: A continuous function. For any sequence of points (xn)n∈N converging to x0, thesequence (f (xn))n∈N converges to f (x0).
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Continuity and Limits of Functions
Continuous Functions.
ExampleLet f : R→ R be given by f (x) = 2x2 + 1 for all x ∈ R. Prove f iscontinuous on R.
Proof.Suppose we have a real-valued sequence (xn) converging to x0 i.e.lim xn = x0. Then we have
lim f (xn) = lim[2x2n + 1] = 2[lim x2
n ] + 1 = 2x20 + 1 = f (x0),
where the second equality follows by application of the limittheorems. We have shown that for any sequence (xn) converging tox0, we have lim f (xn) = f (x0).This proves that f is continuous at eachx0 ∈ R and so f is continuous on R. �
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Continuous Functions.
ExampleLet f : R→ R be given by f (x) = (1/x) sin(1/x2) for x 6= 0 andf (0) = 0. Show that f is discontinuous at 0.
Proof.It is sufficient to find a sequence (xn) converging to 0 such that(f (xn)) does not converge to f (0) = 0. To find such a sequence, wewill rearrange (1/xn) sin(1/x2
n ) = 1/xn where xn → 0. Thus, we wantsin(1/x2
n ) = 1. For this we need 1/x2n = 2nπ + π/2, and it follows that
xn =1√
2nπ + π2.
Then lim xn = 0, while lim f (xn) = lim(1/xn) = +∞ 6= 0. �
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Continuity and Limits of Functions
Continuous Functions.I The sequential definition of continuity implies that the values f (x)
are close to f (x0) when the values x are close to x0. Thefollowing theorem provides an alternative definition of continuity.
Theorem (ε-δ definition of continuity)Let f be a real-valued function whose domain is a subset of R. Then fis continuous at x0 ∈ dom(f ) iff
for each ε > 0 there exists δ > 0 such thatx ∈ dom(f ) and |x−x0| < δ imply |f (x)−f (x0)| < ε.
I If we draw two horizontal lines, no matter how close together, wecan always cut off a vertical strip of the plane by two verticallines in such a way that all that part of the curve which iscontained in the strip lies between the two horizontal lines.
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Continuity and Limits of Functions
Continuous Functions.Epsilon Delta Applet
x
y
f (x0)− εf (x0)
f (x0) + ε
x0 − δ x0 x0 + δFigure: ε-δ definition of continuity
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Continuity and Limits of Functions
Continuous Functions.
ExampleLet f : R→ R be given by f (x) = x2 sin(1/x) for all x 6= 0 andf (0) = 0. Prove f is continuous at 0.
Proof.Let ε > 0. Clearly |f (x)− f (0)| = |f (x)| ≤ x2 for all x . We want this tobe less than ε, so set δ =
√ε. Then |x − 0| < δ implies x2 < δ2 = ε,
and so|x − 0| < δ implies |f (x)− f (0)| < ε.
So f is continuous at 0. �
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Continuous Functions.
ExampleLet f : R→ R be given by f (x) = 2x2 + 1 for all x ∈ R. Then f iscontinuous on R.
Proof.We now use the ε-δ property. Let x0 ∈ R, and let ε > 0. We want toshow that |f (x)− f (x0)| < ε provided |x − x0| is sufficiently small , i.e.less than some δ. First note that
|f (x)− f (x0)| = |2x2 + 1− (2x20 + 1)| = |2x2 − 2x2
0 |= |2(x − x0)(x + x0)| = 2|x − x0||x + x0|.
So, what we need is to find a bound for |x + x0| that does not dependon x .
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Continuity and Limits of Functions
Continuous Functions.
Now, if |x − x0| < 1 then |x | < |x0|+ 1 and hence|x + x0| ≤ |x |+ |x0| < 2|x0|+ 1. So, we have
|f (x)− f (x0)| < 2|x − x0|(2|x0 + 1)
if |x − x0| < 1. To arrange for 2|x − x0|(2|x0 + 1) < ε it is sufficient tohave |x − x0| < ε/[2(2|x0 + 1)] and also |x − x0| < 1. So let
δ = min{
1,ε
2(2|x0|+ 1)
}The working above shows that |x − x0| < δ implies|f (x)− f (x0)| < ε. �
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Continuous Functions.
I We can form new functions from old functions in several ways.
Definition.Let A ⊆ R and let B ⊆ R. Consider two functions f : A→ R andg : B → R and let k ∈ R. We define the functions into R as follows.
I |f | given by |f |(x) = |f (x)| for all x ∈ A;I kf given by (kf )(x) = kf (x) for all x ∈ A;I f + g given by (f + g)(x) = f (x) + g(x) for all x ∈ A ∩ B;I fg given by (fg)(x) = f (x)g(x) for all x ∈ A ∩ B;I f/g given by (f/g)(x) = f (x)/g(x) for all x ∈ A ∩ B such that
g(x) 6= 0. N
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Continuous Functions.
TheoremLet f and g be real-valued functions that are continuous at x0 in Rand let k ∈ R. Then
(i) |f | is continuous at x0;(ii) kf is continuous at x0;
(iii) f + g is continuous at x0;(iv) fg is continuous at x0;(v) f/g is continuous at x0 if g(x0) 6= 0.
TheoremIf f is continuous at x0 and g is continuous at f (x0), then thecomposite function g ◦ f is continuous at x0.
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Continuity and Limits of Functions
Properties of Continuous Functions.
Theorem (Intermediate Value Theorem)If f is a continuous real-valued function on an interval I, then f hasthe intermediate value property on I: Whenever a,b ∈ I, a < b and ylies between f (a) and f (b), there exists at least one x ∈ (a,b) suchthat f (x) = y.
I This theorem can be used to establish that a continuous functionf has a fixed point, i.e. a point x0 ∈ dom(f ) such that f (x0) = x0.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Properties of Continuous Functions.
x
y0
f (b)
y0
f (a)
x1 x2 x3a b[ ]
I
f (x)
Figure: By the intermediate value theorem, for any f (a) < y < f (b), we can find an x such thatf (x) = y . In the case of y0, we can find three i.e. y0 = f (x1) = f (x2) = f (x3).
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Properties of Continuous Functions.ExampleShow that the continuous function f : [0,1]→ [0,1] has a fixed pointx0 ∈ [0,1].
x
y
1
1
y=
x
f (x)
x0
Figure: Fixed point. Kinda obvious?
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Continuity and Limits of Functions
Properties of Continuous Functions.
I We will use the IVT. We will define a new function g on I, andconsider a = 0 and b = 1.
Solution.Consider the function g(x) = f (x)− x , which is also continuous on[0,1] by (iii). Now
g(0) = f (0)− 0 = f (0) ≥ 0
andg(1) = f (1)− 1 ≤ 1− 1 = 0.
So, by the intermediate value theorem, we have that g(x0) = 0 forsome x0 ∈ [0,1]. Then clearly f (x0) = x0. �
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
Definition.Let S be a subset of R, let a be a real number or symbol∞ or −∞that is the limit of some sequence in S, and let L be a real number orsymbol∞ or −∞. We write limx→aS f (x) = L if
I f is a function defined on S, andI for every sequence (xn) in S with limit a, we have
limn→∞ f (xn) = L. N
I The expression “limx→aS f (x)” is read “limit, as x tends to a alongS, of f (x)”.
I Using this definition, we see that a function f is continuous at a indom(f ) = S iff limx→aS f (x) = f (a).
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.Definition.For a function f and a ∈ R we write
(i) limx→a f (x) = L provided limx→aS f (x) = L for some set S = J\awhere J is an open interval containing a. limx→a f (x) is calledthe (two-sided) limit of f at a.
(ii) limx→a+ f (x) = L provided limx→aS f (x) = L for some openinterval S = (a,b). limx→a+ f (x) is called the right-hand limit of fat a.
(iii) limx→a− f (x) = L provided limx→aS f (x) = L for some openinterval S = (c,a). limx→a− f (x) is called the left-hand limit of fat a.
(iv) limx→∞ f (x) = L provided limx→∞S f (x) = L for some intervalS = (c,∞).
(v) limx→−∞ f (x) = L provided limx→−∞S f (x) = L for some intervalS = (−∞,b). N
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Continuity and Limits of Functions
Limits of Functions.
x
y
t
s
a
f (x)
Figure: The right hand limit at a is t , while the left hand limit is s.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.Example
(1) We have limx→4 x3 = 64 and limx→2(1/x) = 1/2 because thefunctions x3 and 1/x are continuous at 4 and 2 respectively. Onecan easily show that limx→0+(1/x) = +∞ andlimx→0−(1/x) = −∞. It follows that limx→0(1/x) does not exist(see theorem 7).
(2) Consider
limx→2
[x2 − 4x − 2
].
The function we are finding the limit of is not defined at x = 2. Wecan rewrite the function as
x2 − 4x − 2
=(x − 2)(x + 2)
x − 2= x + 2 for x 6= 2.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.Now we can see that
limx→2
[x2 − 4x − 2
]= lim
x→2(x + 2) = 4.
It is important to note that the functions given by (x2 − 4)/(x − 2)and (x + 2) are not identical. The domain of the first is(−∞,2) ∪ (2,+∞), while the domain of the second is R.
(3) Consider
limx→1
[√x − 1
x − 1
].
To find the limit, we multiply the numerator and denominator ofthe function by
√x + 1 to get
√x − 1
x − 1=
x − 1(x − 1)(
√x + 1)
=1√
x + 1for x 6= 1.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
Now we can see that
limx→1
[√x − 1
x − 1
]= lim
x→1
[1√
x + 1
]=
12.
This in fact shows that if h(x) =√
x then h′(1) = 1/2 (as you willsee when we look at differentiation).
(4) Let f be a real-valued function given by f (x) = 1/(x − 2)3 for allx 6= 2. Then
I limx→+∞ f (x) = 0,I limx→−∞ f (x) = 0,
I limx→2+ f (x) = +∞,I limx→2− f (x) = −∞.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
Proof.We will show that limx→∞ f (x) = 0. By definition, it is enough to showthat limx→∞S f (x) = 0 for S = (2,+∞). So consider any sequence(xn) with xn ∈ S for all n ∈ N, such that limn→+∞ xn = +∞ and showthat
limn→+∞
f (xn) = limn→+∞
[1
(xn − 2)3
]= 0.
To prove the above assertion we could use our limit theorems.Instead we will prove it directly.Let ε > 0. We need to find an N such that for n > N, we have|(xn − 2)−3| < ε. This inequality can be rearranged asε−1 < |(xn − 2)3| or ε−
13 < |xn − 2|. We need xn > ε−
13 + 2, for the
previous inequality to be satsified.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
Now we use the fact that limn→+∞ xn = +∞. By definition of aninfinite limit of a sequence, for any M > 0 we can find an N such thatn > N implies xn > M. Thus, if we set M = ε−
13 + 2, there exists an N
such thatn > N implies xn > ε−
13 + 2
Then reversing the steps above, we find that
n > N implies |(xn − 2)−3| < ε.
This shows that limn→+∞ f (xn) = 0 for any sequence (xn) in S suchthat limn→+∞ xn = +∞ and the result follows by definition. �
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
I The following result allows us to avoid sequences and provideε-δ definitions of a function’s limits.
TheoremLet f be a function defined on a subset S of R, let a be a real numberthat is the limit of some sequence in S, and let L be a real number.Then limx→as f (x) = L iff
for each ε > 0 there exists δ > 0 such thatx ∈ S, |x − a| < δ imply |f (x)− L| < ε.
I This theorem has a number of corollaries, which are listed in thenext theorem. These give us alternative definitions for the limit ofa function, its lateral limits, and its limits at infinity.
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Continuity and Limits of Functions
Limits of Functions.Theorem
(i) Let f be a function defined on J\{a} for some open interval Jcontaining a, and let L be a real number. Then limx→a f (x) = L iff
for each ε > 0 there exists δ > 0 such that0 < |x − a| < δ implies |f (x)− L| < ε.
(ii) Let f be a function defined on some interval (a,b), and let L be areal number. Then limx→a+ f (x) = L iff
for each ε > 0 there exists δ > 0 such thata < x < a + δ implies |f (x)− L| < ε.
(iii) Let f be a function defined on some interval (c,a), and let L be areal number. Then limx→a− f (x) = L iff
for each ε > 0 there exists δ > 0 such thata− δ < x < a implies |f (x)− L| < ε.
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Continuity and Limits of Functions
Limits of Functions.
x
y
t
s + ε
ss − ε
a + δaa− δ
f (x)
Figure: The (two-sided) limit at a is s, while f (a) = t . (This function is not continuous at a.)
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Continuity and Limits of Functions
Limits of Functions.
x
y
t + ε
tt − ε
s + ε
ss − ε
a + δaa− δ
f (x)
Figure: The right hand limit at a is t , while the left hand limit is s.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
(iv) Let f be a function defined on some interval (c,+∞), and let Lbe a real number. Then limx→+∞ f (x) = L iff
for each ε > 0 there exists δ < +∞ such thatx > δ implies |f (x)− L| < ε.
(v) Let f be a function defined on some interval (−∞,b), and let Lbe a real number. Then limx→−∞ f (x) = L iff
for each ε > 0 there exists δ > −∞ such thatx < δ implies |f (x)− L| < ε.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
x
y
f (x)
ε
−ε
1ε
−1ε
Figure: The function f : R \ {0} → R is given by f (x) = 1/x for all x ∈ R \ {0}. The limits at+∞ and −∞ are both 0.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
ExampleLet f : R \ {0} → R be given by f (x) = 1/x for all x ∈ R \ {0} (this isthe function drawn in the previous figure). Thenlimx→+∞ = limx→−∞ = 0.
Proof.We will prove the first limit. First note that the function is defined onsome interval (c,+∞) – we could take any c > 0. Also the limit L = 0is a real number. So we use definition iv. Let ε > 0. We need to find afinite δ such that x > δ implies |1/x | < ε. If we take δ = 1/ε, then thedesired condition is satisfied. �
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.(vi) Let f be a function defined on J \ {a}. Then limx→a f (x) =∞ iff
for each M > 0 there exists δ > 0 such that0 < |x − a| < δ implies f (x) > M.
(vii) Let f be a function defined on J \ {a}. Then limx→a f (x) = −∞ iff
for each M < 0 there exists δ > 0 such that0 < |x − a| < δ implies f (x) < M.
I We can also mix the previous limit concepts, and define thefollowing limits.
I limx→a+ f (x) = +∞,I limx→a+ f (x) = −∞,I limx→a− f (x) = +∞,I limx→a− f (x) = −∞,
I limx→+∞ f (x) = +∞,I limx→+∞ f (x) = −∞,I limx→−∞ f (x) = +∞,I limx→−∞ f (x) = +∞.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
x
y
f (x)M
−M
1M
− 1M
Figure: The function f : R \ {0} → R is given by f (x) = 1/x for all x ∈ R \ {0}. Herelimx→0+ f (x) = +∞, while limx→0− f (x) = −∞. The two-sided limit at a does not exist. Itcannot be +∞ because there is no δ such that x ∈ (−δ,+δ) implies f (x) > M.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
I The next result tells us that if the left-hand and right-hand limitsat a of a function exist and are equal, then the two-sided limit at aexists and is equal to these two limits.
I It also says the converse is also true – if the limit exists, thenboth the left-hand and right-hand limits exist and are equal to thetwo-sided limit.
TheoremLet f be a function defined on J\{a} for some open interval Jcontaining a. Then limx→a f (x) exists iff the limits limx→a+ f (x) andlimx→a− f (x) both exist and are equal, in which case all three limitsare equal.
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.
TheoremLet f1 and f2 be functions for which the limits L1 = limx→aS f1(x) andL2 = limx→aS f2(x) exist and are finite. Then
(i) limx→aS (f1 + f2)(x) = L1 + L2;(ii) limx→aS (f1f2)(x) = L1L2;(iii) limx→aS (f1/f2)(x) = L1/L2 if L2 6= 0 and f2(x) 6= 0 for x ∈ S.
Proof of (iii).blah �
Scott McCracken MTAEA – Continuity and Limits of Functions
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Continuity and Limits of Functions
Limits of Functions.I The following result provides conditions under which we can
change the order of taking the limit and applying the function g.
TheoremLet f be a function for which the limit L = limx→aS f (x) exists and isfinite. If g is a function defined on {f (x) | x ∈ S} ∪ {L} that iscontinuous at L, then limx→aS g ◦ f (x) exists and equals g(L).
ExampleCompute limx→1
√x3 + 1.
Proof.Here the function g : R+ → R, given by g(x) =
√x for all x ∈ R+ is
continuous. Thus
limx→1
√x3 + 1 =
√limx→1
x3 + 1 =√
1 + 1 =√
2. �
Scott McCracken MTAEA – Continuity and Limits of Functions