Lesson 4: Calculating Limits (handout)

Section 1.4Calculating Limits

V63.0121.021, Calculus I

New York University

September 15, 2010

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

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I First written homework duetoday (put it in theenvelope) Remember to putyour lecture and recitationsection numbers on yourpaper

V63.0121.021, 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 ourselves only tospeak in terms of what we personally have produced.


Notes

Notes

Notes

1

Section 1.4 : Calculating LimitsV63.0121.021, Calculus I September 15, 2010

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 Theorem todemonstrate 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

Definition

We writelimx→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 as welike) by taking x to be sufficiently close to a (on either side of a) but notequal to a.


Notes

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2


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 withinthat tolerance of a (not counting a itself) are taken to values ywithin the error level of L. If Dana cannot, Emerson wins and thelimit cannot be L.

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


The error-tolerance game

This tolerance is too bigStill too bigThis looks goodSo does this

a

L

I To be legit, the part of the graph inside the blue (vertical) strip mustalso 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 graph in-side blue is notinside green

Part of graph in-side blue is notinside green

I So limx→0

|x |x

does not exist.


Notes

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3


Limit FAIL: unboundedness

x

y

0

L?

The graph escapes thegreen, so no good

Even worse!lim

x→0+

1

xdoes not exist

because the function isunbounded 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 close tozero where f (x) = y . So lim

x→0f (x) cannot exist.


Outline


Basic Limits





Notes

Notes

Notes

4


Really basic limits

Fact

Let c be a constant and a a real number.

(i) limx→a

x = 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!


Notes

Notes

Notes

5


Really basic limits

Fact

Let c be a constant and a a real number.

(i) limx→a

x = a

(ii) limx→a

c = c

Proof.

The first is tautological, the second is trivial.


Outline


Basic Limits





Limits and arithmetic

Fact

Suppose limx→a

f (x) = L and limx→a

g(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)


Notes

Notes

Notes

6



Fact

Suppose limx→a



1. limx→a


2. limx→a


3. limx→a


4. limx→a



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 thefact 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.



Fact

Suppose limx→a



1. limx→a


2. limx→a


3. limx→a


4. limx→a



Notes

Notes

Notes

7



Fact

Suppose limx→a



1. limx→a


2. limx→a


3. limx→a


4. limx→a



Limits and arithmetic II

Fact (Continued)

5. limx→a

f (x)

g(x)=

L

M, if M 6= 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 assume

that limx→a

f (x) > 0)


Caution!

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

g(x) 6= 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


Notes

Notes

Notes

8


Limits and arithmetic II

Fact (Continued)

5. limx→a

f (x)

g(x)=

L

M, if M 6= 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 assume

that limx→a

f (x) > 0)


Applying the limit laws

Example

Find limx→3

(x2 + 2x + 4

).

Solution

By applying the limit laws repeatedly:

limx→3

(x2 + 2x + 4

)= lim

x→3

(x2)

+ limx→3

(2x) + limx→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 + 4

x3 + 11

Solution

The answer is19

38=

1

2.


Notes

Notes

Notes

9


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)

Theorem

If f (x) = g(x) when x 6= a, and limx→a

g(x) = L, then limx→a

f (x) = L.

Example

Find limx→−1

x2 + 2x + 1

x + 1, if it exists.

Solution

Sincex2 + 2x + 1

x + 1= x + 1 whenever x 6= −1, and since lim

x→−1x + 1 = 0,

we have limx→−1

x2 + 2x + 1

x + 1= 0.


Notes

Notes

Notes

10


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 boundarypoint

Example

Let

f (x) =

{x2 x ≥ 0

−x x < 0

Does limx→0

f (x) exist?

Solution

We havelim

x→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.V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 15, 2010 32 / 45

Finding limits by algebraic manipulations

Example

Find limx→4

√x − 2

x − 4.

Solution

Write the denominator as x − 4 =√x2 − 4 = (

√x − 2)(

√x + 2). So

limx→4

√x − 2

x − 4= lim

x→4

√x − 2

(√x − 2)(

√x + 2)

= limx→4

1√x + 2

=1

4


Notes

Notes

Notes

11


Your turn

Example

Let

f (x) =

{1− x2 x ≥ 1

2x x < 1

Find limx→1

f (x) if it exists. 1

Solution

We 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.V63.0121.021, 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 verbs

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

f (x) DNE and limx→a

g(x) DNE,

then limx→a

(f (x) + g(x)) DNE.”


Two More Important Limit Theorems

Theorem

If 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.


Notes

Notes

Notes

12


Using the Squeeze Theorem

We can use the Squeeze Theorem to replace complicated expressions withsimple ones when taking the limit.

Example

Show that limx→0

x2 sin(πx

)= 0.

Solution

We 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





Notes

Notes

Notes

13



Theorem

The following two limits hold:

I limθ→0

sin θ

θ= 1

I limθ→0

cos θ − 1

θ= 0


Proof of the Sine Limit

Proof.

θsin 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 θ

θ≥ cos θ

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


Proof of the Cosine Limit

Proof.

1− cos θ

θ=

1− cos θ

θ· 1 + cos θ

1 + cos θ=

1− cos2 θ

θ(1 + cos θ)

=sin2 θ

θ(1 + cos θ)=

sin θ

θ· sin θ

1 + cos θ

So

limθ→0

1− cos θ

θ=

(limθ→0

sin θ

θ

)·(

limθ→0

sin θ

1 + cos θ

)= 1 · 0 = 0.


Notes

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14


Try these

Example

1. limθ→0

tan θ

θ

2. limθ→0

sin 2θ

θ

Answer

1. 1

2. 2


Solutions

1. Use the basic trigonometric limit and the definition of tangent.

limθ→0

tan θ

θ= lim

θ→0

sin θ

θ cos θ= lim

θ→0

sin θ

θ· limθ→0

1

cos θ= 1 · 1

1= 1.

2. Change the variable:

limθ→0

sin 2θ

θ= lim

2θ→0

sin 2θ

2θ · 12= 2 · lim

2θ→0

sin 2θ

2θ= 2 · 1 = 2

OR use a trigonometric identity:

limθ→0

sin 2θ

θ= lim

θ→0

2 sin θ cos θ

θ= 2 · lim

θ→0

sin θ

θ· 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.


Notes

Notes

Notes

15


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