Lesson 4: Calcuating Limits (slides)

..

Sec on 1.4Calcula ng Limits

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

February 2, 2011

Announcements

I First wri en HW due todayI Get-to-know-you survey and photo deadline is February 11

Announcements

I First wri en HW duetoday

I Get-to-know-you surveyand photo deadline isFebruary 11

ObjectivesI Know basic limits like lim

x→ax = a

and limx→a

c = c.I Use the limit laws to computeelementary limits.

I Use algebra to simplify limits.I Understand and state theSqueeze Theorem.

I Use the Squeeze Theorem todemonstrate a limit.

..

Limit

Yoda on teaching course concepts

You must unlearnwhat you havelearned.

In other words, we arebuilding up concepts andallowing ourselves only tospeak in terms of what wepersonally have produced.

OutlineRecall: The concept of limit

Basic Limits

Limit LawsThe direct subs tu on property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits

Heuristic Definition of a LimitDefini onWe write

limx→a

f(x) = L

and say

“the limit of f(x), as x approaches a, equals L”

if we can make the values of f(x) arbitrarily close to L (as close to Las we like) by taking x to be sufficiently close to a (on either side ofa) but not equal to a.

The error-tolerance gameA game between two players (Dana and Emerson) to decide if a limitlimx→a

f(x) exists.

Step 1 Dana proposes L to be the limit.Step 2 Emerson challenges with an “error” level around L.Step 3 Dana chooses a “tolerance” level around a so that points x

within that tolerance of a (not coun ng a itself) are taken tovalues y within the error level of L. If Dana cannot, Emersonwins and the limit cannot be L.

Step 4 If Dana’s move is a good one, Emerson can challenge againor give up. If Emerson gives up, Dana wins and the limit is L.

The error-tolerance game

.

.

This tolerance is too big

.

S ll too big

.

This looks good

.

So does this

.a

.

L

I To be legit, the part of the graph inside the blue (ver cal) stripmust also be inside the green (horizontal) strip.

I Even if Emerson shrinks the error, Dana can s ll move.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

...

Part of graphinside blueis not insidegreen

.


I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

..

.


.


I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

Limit FAIL: unboundedness

.. x.

y

.0..

L?

.

The graph escapesthe green, so nogood

.

Even worse!

.

limx→0+

1xdoes not exist be-

cause the func on is un-bounded near 0

Limit EPIC FAILHere is a graph of the func on f(x) = sin

(πx

):

.. x.

y

..

−1

..

1

For every y in [−1, 1], there are infinitely many points x arbitrarilyclose to zero where f(x) = y. So lim

x→0f(x) cannot exist.


Basic Limits




Really basic limits

FactLet c be a constant and a a real number.(i) lim

x→ax = a

(ii) limx→a

c = c

Proof.The first is tautological, the second is trivial.

ET game for f(x) = x

.. x.

y

..a

..

aI Se ng error equal totolerance works!


.. x.

y

..a

..

a

I Se ng error equal totolerance works!


.. x.

y

..a

..

aI Se ng error equal totolerance works!

ET game for f(x) = c

.

. x.

y

..a

..

cI any tolerance works!


.. x.

y

..a

..



.. x.

y

..a

..

c

I any tolerance works!


.. x.

y

..a

..


Really basic limits

FactLet c be a constant and a a real number.(i) lim

x→ax = a

(ii) limx→a

c = c

Proof.The first is tautological, the second is trivial.


Basic Limits




Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M

(errors add)

2. limx→a

[f(x)− g(x)] = L−M

(combina on of adding and scaling)

3. limx→a

[cf(x)] = cL

(error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)


FactSuppose lim



1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M


3. limx→a

[cf(x)] = cL

(error scales)

4. limx→a

[f(x)g(x)] = L ·M



FactSuppose lim



1. limx→a


2. limx→a

[f(x)− g(x)] = L−M


3. limx→a

[cf(x)] = cL (error scales)

4. limx→a

[f(x)g(x)] = L ·M


Justification of the scaling lawI errors scale: If f(x) is e away from L, then

(c · f(x)− c · L) = c · (f(x)− L) = c · e

That is, (c · f)(x) is c · e away from cL,I So if Emerson gives us an error of 1 (for instance), Dana can usethe fact that lim

x→af(x) = L to find a tolerance for f and g

corresponding to the error 1/c.I Dana wins the round.


FactSuppose lim



1. limx→a


2. limx→a

[f(x)− g(x)] = L−M (combina on of adding and scaling)

3. limx→a


4. limx→a

[f(x)g(x)] = L ·M



FactSuppose lim



1. limx→a


2. limx→a

[f(x)− g(x)] = L−M (combina on of adding and scaling)

3. limx→a


4. limx→a

[f(x)g(x)] = L ·M (more complicated, but doable)

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an

(follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Caution!I The quo ent rule for limits says that if lim

x→ag(x) ̸= 0, then

limx→a

f(x)g(x)

=limx→a f(x)limx→a g(x)

I It does NOT say that if limx→a

g(x) = 0, then

limx→a

f(x)g(x)

does not exist

I In fact, limits of quo ents where numerator and denominatorboth tend to 0 are exactly where the magic happens.


5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n


7. limx→a

xn = an

(follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a


x→af(x) > 0)


5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n


7. limx→a

xn = an (follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a


x→af(x) > 0)

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)

= limx→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.


Find limx→3

(x2 + 2x+ 4

).


limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.


Find limx→3

(x2 + 2x+ 4

).


limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4

= 9+ 6+ 4 = 19.


Find limx→3

(x2 + 2x+ 4

).


limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.

Your turnExample

Find limx→3

x2 + 2x+ 4x3 + 11

Solu on

The answer is1938

=12.

Direct Substitution Property

As a direct consequence of the limit laws and the really basic limitswe have:Theorem (The Direct Subs tu on Property)

If f is a polynomial or a ra onal func on and a is in the domain of f,then

limx→a

f(x) = f(a)


Basic Limits




Limits do not see the point!

(in a good way)

TheoremIf f(x) = g(x) when x ̸= a, and lim

x→ag(x) = L, then lim

x→af(x) = L.

Example of the MTP principleExample

Find limx→−1

x2 + 2x+ 1x+ 1

, if it exists.

Solu on

Sincex2 + 2x+ 1

x+ 1= x+ 1 whenever x ̸= −1, and since

limx→−1

x+ 1 = 0, we have limx→−1

x2 + 2x+ 1x+ 1

= 0.

ET game for f(x) =x2 + 2x + 1

x + 1

.. x.

y

...−1

I Even if f(−1) were something else, it would not effect the limit.

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

.

.


Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0


f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

..


Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0


f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

.

.

Finding limits by algebraExample

Find limx→4

√x− 2x− 4

.

Solu onWrite the denominator as x− 4 =

√x2 − 4 = (

√x− 2)(

√x+ 2). So

limx→4

√x− 2x− 4

= limx→4

√x− 2

(√x− 2)(

√x+ 2)

= limx→4

1√x+ 2

=14


Find limx→4

√x− 2x− 4

.


√x2 − 4 = (

√x− 2)(

√x+ 2).

So

limx→4

√x− 2x− 4

= limx→4

√x− 2

(√x− 2)(

√x+ 2)

= limx→4

1√x+ 2

=14


Find limx→4

√x− 2x− 4

.


√x2 − 4 = (

√x− 2)(

√x+ 2). So

limx→4

√x− 2x− 4

= limx→4

√x− 2

(√x− 2)(

√x+ 2)

= limx→4

1√x+ 2

=14

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2

I The le - and right-hand limits disagree, so thelimit does not exist.

.

..1

..

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2


...1

.

.

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2


...1

..

A message fromthe Mathematical Grammar Police

Please do not say “limx→a

f(x) = DNE.” Does not compute.

I Too many verbsI Leads to FALSE limit laws like “If lim

x→af(x) DNE and lim

x→ag(x) DNE,

then limx→a

(f(x) + g(x)) DNE.”



f(x) = DNE.” Does not compute.I Too many verbs

I Leads to FALSE limit laws like “If limx→a

f(x) DNE and limx→a

g(x) DNE,then lim

x→a(f(x) + g(x)) DNE.”



f(x) = DNE.” Does not compute.I Too many verbsI Leads to FALSE limit laws like “If lim

x→af(x) DNE and lim

x→ag(x) DNE,

then limx→a

(f(x) + g(x)) DNE.”

Two Important Limit Theorems

TheoremIf f(x) ≤ g(x) when x is near a(except possibly at a), then

limx→a

f(x) ≤ limx→a

g(x)

(as usual, provided these limitsexist).

Theorem (The Squeeze/Sandwich/ Pinching Theorem)

If f(x) ≤ g(x) ≤ h(x) when x isnear a (as usual, exceptpossibly at a), and

limx→a

f(x) = limx→a

h(x) = L,

then limx→a

g(x) = L.

Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicatedexpressions with simple ones when taking the limit.

Example

Show that limx→0

x2 sin(πx

)= 0.

Solu onWe have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2

The le and right sides go to zero as x → 0.

Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicatedexpressions with simple ones when taking the limit.Example

Show that limx→0

x2 sin(πx

)= 0.

Solu onWe have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2

The le and right sides go to zero as x → 0.

Illustrating the Squeeze Theorem

.. x.

y

.

h(x) = x2

.

f(x) = −x2

.

g(x) = x2 sin(πx

)


Basic Limits




Two trigonometric limits

TheoremThe following two limits hold:

I limθ→0

sin θθ

= 1

I limθ→0

cos θ − 1θ

= 0

Proof of the Sine LimitProof.

.. θ

.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce

sin θ ≤

θ

≤ 2 tanθ

2≤ tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θI Take reciprocals: 1 ≥ sin θ

θ≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.


.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ

≤ 2 tanθ

2≤ tan θ


sin θ≤ 1


θ≥ cos θ



.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ

≤ 2 tanθ

2≤

tan θ


sin θ≤ 1


θ≥ cos θ



.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ ≤ 2 tanθ

2≤ tan θ


sin θ≤ 1


θ≥ cos θ



.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1


2≤ tan θ


sin θ≤ 1

cos θ

I Take reciprocals: 1 ≥ sin θθ

≥ cos θ



.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1


2≤ tan θ


sin θ≤ 1


θ≥ cos θ


Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.


1− cos θθ

=1− cos θ



=sin2 θ

θ(1+ cos θ)

=sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.


1− cos θθ

=1− cos θ



=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.


1− cos θθ

=1− cos θ



=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)

= 1 · 02= 0.


1− cos θθ

=1− cos θ



=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.

Try theseExample

1. limθ→0

tan θθ

2. limθ→0

sin 2θθ

Answer

1. 12. 2

Solutions1. Use the basic trigonometric limit and the defini on of tangent.

limθ→0

tan θθ

= limθ→0

sin θθ cos θ

= limθ→0

sin θθ

· limθ→0

1cos θ

= 1 · 11= 1.

2. Change the variable:

limθ→0

sin 2θθ

= lim2θ→0

sin 2θ2θ · 1

2= 2 · lim

2θ→0

sin 2θ2θ

= 2 · 1 = 2

OR use a trigonometric iden ty:

limθ→0

sin 2θθ

= limθ→0

2 sin θ cos θθ

= 2·limθ→0

sin θθ

·limθ→0

cos θ = 2·1·1 = 2

Summary

I The limit laws allow us tocompute limitsreasonably.

I BUT we cannot make upextra laws otherwise weget into trouble.

.. x.

y

Lesson 4: Calcuating Limits (slides)

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Transcript of Lesson 4: Calcuating Limits (slides)