lesson 4: calculating limits (section 41 slides)

Section 1.4Calculating Limits

V63.0121.041, Calculus I

New York University

September 15, 2010

Announcements

I First written homework due today (put it in the envelope)Remember to put your lecture and recitation section numbers onyour paper

. . . . . .

. . . . . .

Announcements

I First written homework duetoday (put it in theenvelope) Remember toput your lecture andrecitation section numberson your paper

V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 2 / 45

. . . . . .

Yoda on teaching a concepts course

“You must unlearn what you have learned.”

In other words, we are building up concepts and allowing ourselvesonly to speak in terms of what we personally have produced.


. . . . . .

Objectives

I Know basic limits likelimx→a

x = a and limx→a

c = c.

I Use the limit laws tocompute elementary limits.

I Use algebra to simplifylimits.

I Understand and state theSqueeze Theorem.

I Use the Squeeze Theoremto demonstrate a limit.


. . . . . .

Outline

Recall: The concept of limit

Basic Limits

Limit LawsThe direct substitution property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits


. . . . . .

Heuristic Definition of a Limit

DefinitionWe 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 L aswe like) by taking x to be sufficiently close to a (on either side of a) butnot equal to a.


. . . . . .

The error-tolerance game

A 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 counting 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 again orgive up. If Emerson gives up, Dana wins and the limit is L.


. . . . . .

The error-tolerance game

.

.This tolerance is too big.Still too big.This looks good.So does this

.a

.L

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

I If Emerson shrinks the error, Dana can still win.


. . . . . .

Limit FAIL: Jump

. .x

.y

..−1

..1 .

..Part of graphinside blue is notinside green

.Part of graphinside blue is notinside green

I So limx→0

|x|x

does not exist.


. . . . . .

Limit FAIL: Jump

. .x

.y

..−1

..1 .

.



I So limx→0

|x|x

does not exist.


. . . . . .

Limit FAIL: unboundedness

. .x

.y

.0

..L?

.The graph escapesthe green, so no good.Even worse!

.lim

x→0+1xdoes not exist

because the functionis unbounded near 0


. . . . . .

Limit EPIC FAIL

Here is a graph of the function f(x) = sin(πx

):

. .x

.y

..−1

..1

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

x→0f(x) cannot exist.


. . . . . .

Outline


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

..a

I Setting error equal to tolerance works!


. . . . . .

ET game for f(x) = c

.

.x

.y

..a

..c

I any tolerance works!


. . . . . .

ET game for f(x) = c

. .x

.y

..a

..c

I any tolerance works!


. . . . . .

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.


. . . . . .

Outline


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

(combination 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



. . . . . .


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 law

I 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 use

the fact that limx→a

f(x) = L to find a tolerance for f and gcorresponding to the error 1/c.

I Dana wins the round.


. . . . . .


FactSuppose lim



1. limx→a


2. limx→a

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

3. limx→a


4. limx→a

[f(x)g(x)] = L ·M



. . . . . .


FactSuppose lim



1. limx→a


2. limx→a


3. limx→a


4. limx→a

[f(x)g(x)] = L ·M



. . . . . .


FactSuppose lim



1. limx→a


2. limx→a


3. limx→a


4. limx→a

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


. . . . . .

Limits and arithmetic II

Fact (Continued)

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 additionally assumethat lim

x→af(x) > 0)


. . . . . .

Caution!

I The quotient rule for limits says that if limx→a

g(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

In fact, limits of quotients where numerator and denominator bothtend to 0 are exactly where the magic happens.

I more about this later


. . . . . .


Fact (Continued)

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)


. . . . . .


Fact (Continued)

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 laws

Example

Find limx→3

(x2 + 2x+ 4

).

SolutionBy 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.


. . . . . .


Example

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.


. . . . . .


Example

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.


. . . . . .


Example

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 turn

Example

Find limx→3

x2 + 2x+ 4x3 + 11

Solution

The answer is1938

=12.


. . . . . .

Direct Substitution Property

Theorem (The Direct Substitution Property)

If f is a polynomial or a rational function and a is in the domain of f, then

limx→a

f(x) = f(a)


. . . . . .

Outline


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

Find limx→−1

x2 + 2x+ 1x+ 1

, if it exists.

Solution

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 function defined piecewise at a boundary

point

Example

Let

f(x) =

{x2 x ≥ 0−x x < 0

Does limx→0

f(x) exist?

.

.

SolutionWe have

limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

Likewise:lim

x→0−f(x) = lim

x→0−−x = −0 = 0

So limx→0

f(x) = 0.


. . . . . .


point

Example

Let

f(x) =

{x2 x ≥ 0−x x < 0

Does limx→0

f(x) exist?

..

SolutionWe have

limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

Likewise:lim

x→0−f(x) = lim

x→0−−x = −0 = 0

So limx→0

f(x) = 0.


. . . . . .


point

Example

Let

f(x) =

{x2 x ≥ 0−x x < 0

Does limx→0

f(x) exist?

.

.

SolutionWe have

limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

Likewise:lim

x→0−f(x) = lim

x→0−−x = −0 = 0

So limx→0

f(x) = 0.


. . . . . .


point

Example

Let

f(x) =

{x2 x ≥ 0−x x < 0

Does limx→0

f(x) exist?

.

.

SolutionWe have

limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

Likewise:lim

x→0−f(x) = lim

x→0−−x = −0 = 0

So limx→0

f(x) = 0.V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45

. . . . . .

Finding limits by algebraic manipulations

Example

Find limx→4

√x− 2x− 4

.

SolutionWrite 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


. . . . . .


Example

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


. . . . . .


Example

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 turn

Example

Let

f(x) =

{1− x2 x ≥ 12x x < 1

Find limx→1

f(x) if it exists.

.

..1

.

.

SolutionWe have

limx→1+

f(x) = limx→1+

(1− x2

)DSP= 0

limx→1−

f(x) = limx→1−

(2x) DSP= 2

The left- and right-hand limits disagree, so the limit does not exist.


. . . . . .

Your turn

Example

Let

f(x) =

{1− x2 x ≥ 12x x < 1

Find limx→1

f(x) if it exists.

. ..1

.

.

SolutionWe have

limx→1+

f(x) = limx→1+

(1− x2

)DSP= 0

limx→1−

f(x) = limx→1−

(2x) DSP= 2

The left- and right-hand limits disagree, so the limit does not exist.


. . . . . .

Your turn

Example

Let

f(x) =

{1− x2 x ≥ 12x x < 1

Find limx→1

f(x) if it exists.

. ..1

.

.

SolutionWe have

limx→1+

f(x) = limx→1+

(1− x2

)DSP= 0

limx→1−

f(x) = limx→1−

(2x) DSP= 2

The left- and right-hand limits disagree, so the limit does not exist.V63.0121.041, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 34 / 45

. . . . . .

A message from the 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.”


. . . . . .




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.”


. . . . . .




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 More 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 limits exist).

Theorem (The Squeeze/Sandwich/Pinching Theorem)

If f(x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possibly at a),and

limx→a

f(x) = limx→a

h(x) = L,

thenlimx→a

g(x) = L.


. . . . . .

Using the Squeeze Theorem

We can use the Squeeze Theorem to replace complicated expressionswith simple ones when taking the limit.

Example

Show that limx→0

x2 sin(πx

)= 0.

SolutionWe have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2

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


. . . . . .

Illustration of the Squeeze Theorem

. .x

.y .h(x) = x2

.f(x) = −x2

.g(x) = x2 sin(πx

)


. . . . . .

Outline


Basic Limits





. . . . . .


TheoremThe following two limits hold:

I limθ→0

sin θθ

= 1

I limθ→0

cos θ − 1θ

= 0


. . . . . .

Proof of the Sine Limit

Proof.

. .θ

.sin θ

.cos θ

.θ

.tan θ

.−1 .1

Notice

sin θ ≤

θ

≤ 2 tanθ

2≤ tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θθ

≥ cos θ

As θ → 0, the left and right sides tend to 1. So, then, must the middleexpression.


. . . . . .


Proof.

. .θ.sin θ

.cos θ

.θ

.tan θ

.−1 .1

Notice

sin θ ≤ θ

≤ 2 tanθ

2≤ tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θθ

≥ cos θ



. . . . . .


Proof.

. .θ.sin θ

.cos θ

.θ .tan θ

.−1 .1

Notice

sin θ ≤ θ

≤ 2 tanθ

2≤

tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θθ

≥ cos θ



. . . . . .


Proof.

. .θ.sin θ

.cos θ

.θ .tan θ

.−1 .1

Notice

sin θ ≤ θ ≤ 2 tanθ

2≤ tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θθ

≥ cos θ



. . . . . .

Proof of the Cosine Limit

Proof.

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 = 0.


. . . . . .

Try these

Example

1. limθ→0

tan θθ

2. limθ→0

sin 2θθ

Answer

1. 12. 2


. . . . . .

Solutions

1. Use the basic trigonometric limit and the definition 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 identity:

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 to compute limits reasonably.I BUT we cannot make up extra laws otherwise we get into trouble.


lesson 4: calculating limits (section 41 slides)

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