lesson 4: calculating limits (section 21 slides)

Section 1.4Calculating Limits

V63.0121.021, Calculus I

New York University

September 16, 2010

Announcements

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

Announcements

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 16, 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.

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.

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

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

Part of graph in-side blue is notinside green

I So limx→0

does not exist.

Limit FAIL: Jump

I So limx→0

does not exist.

Limit FAIL: Jump

I So limx→0

does not exist.

Limit FAIL: unboundedness

The graph escapes thegreen, so no good

Even worse!

limx→0+

xdoes not exist

because the function isunbounded near 0

Limit EPIC FAIL

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

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

Really basic limits

Let c be a constant and a a real number.

(i) limx→a

(ii) limx→a

Proof.

The first is tautological, the second is trivial.

Really basic limits

(i) limx→a

(ii) limx→a

Proof.

ET game for f (x) = x

I Setting error equal to tolerance works!

ET game for f (x) = c

I any tolerance works!

Really basic limits

(i) limx→a

(ii) limx→a

Proof.

Outline

Basic Limits

Limits and arithmetic

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)

Suppose limx→a

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

Suppose limx→a

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

Suppose limx→a

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

Suppose limx→a

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 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.

Suppose limx→a

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

Suppose limx→a

1. limx→a

2. limx→a

3. limx→a

4. limx→a

[f (x)g(x)] = L ·M

Suppose limx→a

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

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

limx→a f (x)

limx→a g(x)

I It does NOT say that if limx→a

g(x) = 0, then

limx→a

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

M, if M 6= 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

that limx→a

f (x) > 0)

Fact (Continued)

5. limx→a

M, if M 6= 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

that limx→a

f (x) > 0)

Fact (Continued)

5. limx→a

M, if M 6= 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

that limx→a

f (x) > 0)

Fact (Continued)

5. limx→a

M, if M 6= 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

that limx→a

f (x) > 0)

Fact (Continued)

5. limx→a

M, if M 6= 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

that limx→a

f (x) > 0)

Fact (Continued)

5. limx→a

M, if M 6= 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

f (x) = n

√limx→a

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

= limx→3

+ limx→3

(2x) + limx→3

limx→3

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3 + 4

= 9 + 6 + 4 = 19.

Example

Find limx→3

(x2 + 2x + 4

Solution

limx→3

(x2 + 2x + 4

= limx→3

+ limx→3

(2x) + limx→3

limx→3

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3 + 4

= 9 + 6 + 4 = 19.

Example

Find limx→3

(x2 + 2x + 4

Solution

limx→3

(x2 + 2x + 4

)= lim

+ limx→3

(2x) + limx→3

limx→3

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3 + 4

= 9 + 6 + 4 = 19.

Example

Find limx→3

(x2 + 2x + 4

Solution

limx→3

(x2 + 2x + 4

)= lim

+ limx→3

(2x) + limx→3

limx→3

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3 + 4

= 9 + 6 + 4 = 19.

Example

Find limx→3

(x2 + 2x + 4

Solution

limx→3

(x2 + 2x + 4

)= lim

+ limx→3

(2x) + limx→3

limx→3

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3 + 4

= 9 + 6 + 4 = 19.

Example

Find limx→3

(x2 + 2x + 4

Solution

limx→3

(x2 + 2x + 4

)= lim

+ limx→3

(2x) + limx→3

limx→3

+ 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

Your turn

Example

Find limx→3

x2 + 2x + 4

x3 + 11

Solution

The answer is19

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.

Theorem

f (x) = L.

Example

Find limx→−1

x2 + 2x + 1

Solution

Sincex2 + 2x + 1

x→−1x + 1 = 0,

we have limx→−1

x2 + 2x + 1

x + 1= 0.

Theorem

f (x) = L.

Example

Find limx→−1

x2 + 2x + 1

Solution

Sincex2 + 2x + 1

x→−1x + 1 = 0,

we have limx→−1

x2 + 2x + 1

x + 1= 0.

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

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

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

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.

Example

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.

Example

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.

Example

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.

Example

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.

Example

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.

Example

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 16, 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 − 2

(√x − 2)(

√x + 2)

= limx→4

1√x + 2

Example

Find limx→4

√x − 2

x − 4.

Solution

√x − 2)(

√x + 2).

limx→4

√x − 2

x − 4= lim

√x − 2

(√x − 2)(

√x + 2)

= limx→4

1√x + 2

Example

Find limx→4

√x − 2

x − 4.

Solution

√x − 2)(

√x + 2). So

limx→4

√x − 2

x − 4= lim

√x − 2

(√x − 2)(

√x + 2)

= limx→4

1√x + 2

Your turn

Example

f (x) =

{1− x2 x ≥ 1

2x x < 1

Find limx→1

f (x) if it exists.

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.

Your turn

Example

f (x) =

{1− x2 x ≥ 1

2x x < 1

Find limx→1

f (x) if it exists.

Solution

We have

limx→1+

f (x) = limx→1+

(1− x2

) DSP= 0

limx→1−

f (x) = limx→1−

(2x)DSP= 2

Your turn

Example

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

Your turn

Example

f (x) =

{1− x2 x ≥ 1

2x x < 1

Find limx→1

Solution

We have

limx→1+

f (x) = limx→1+

(1− x2

) DSP= 0

limx→1−

f (x) = limx→1−

(2x)DSP= 2

Your turn

Example

f (x) =

{1− x2 x ≥ 1

2x x < 1

Find limx→1

Solution

We have

limx→1+

f (x) = limx→1+

(1− x2

) DSP= 0

limx→1−

f (x) = limx→1−

(2x)DSP= 2

Your turn

Example

f (x) =

{1− x2 x ≥ 1

2x x < 1

Find limx→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 16, 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.”

I Too many verbs

g(x) DNE,

then limx→a

(f (x) + g(x)) DNE.”

I Too many verbs

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

(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 expressions withsimple ones when taking the limit.

Example

Show that limx→0

x2 sin(πx

Solution

We have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

)≤ x2

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

Example

Show that limx→0

x2 sin(πx

Solution

We have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

)≤ x2

Example

Show that limx→0

x2 sin(πx

Solution

We have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

)≤ x2

Illustration of the Squeeze Theorem

y h(x) = x2

f (x) = −x2

g(x) = x2 sin(πx

y h(x) = x2

f (x) = −x2

g(x) = x2 sin(πx

y h(x) = x2

f (x) = −x2

g(x) = x2 sin(πx

Outline

Basic Limits

Theorem

The following two limits hold:

I limθ→0

sin θ

I limθ→0

cos θ − 1

Proof of the Sine Limit

Proof.

sin θ

cos θ

tan θ

−1 1

Notice

sin 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.

θsin θ

cos 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

Proof.

θsin θ

cos θ

θ tan θ

−1 1

Notice

sin θ ≤ θ

≤ 2 tanθ

tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θ

θ≥ cos θ

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

Proof.

θsin θ

cos θ

θ tan θ

−1 1

Notice

2≤ tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θ

θ≥ cos θ

Proof.

θsin θ

cos θ

θ tan θ

−1 1

Notice

2≤ tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θ

θ≥ cos θ

Proof.

θsin θ

cos θ

θ tan θ

−1 1

Notice

2≤ tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θ

θ≥ cos θ

Proof of the Cosine Limit

Proof.

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

limθ→0

1− cos θ

(limθ→0

sin θ

limθ→0

sin θ

1 + cos θ

)= 1 · 0 = 0.

Try these

Example

1. limθ→0

tan θ

2. limθ→0

sin 2θ

Answer

Try these

Example

1. limθ→0

tan θ

2. limθ→0

sin 2θ

Answer

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

cos θ= 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

Solutions

limθ→0

tan θ

θ= lim

θ→0

sin θ

θ cos θ= lim

θ→0

sin θ

θ· limθ→0

cos θ= 1 · 1

limθ→0

sin 2θ

θ= lim

2θ→0

sin 2θ

2θ · 12= 2 · lim

2θ→0

sin 2θ

2θ= 2 · 1 = 2

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

Solutions

limθ→0

tan θ

θ= lim

θ→0

sin θ

θ cos θ= lim

θ→0

sin θ

θ· limθ→0

cos θ= 1 · 1

limθ→0

sin 2θ

θ= lim

2θ→0

sin 2θ

2θ · 12= 2 · lim

2θ→0

sin 2θ

2θ= 2 · 1 = 2

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.

lesson 4: calculating limits (section 21 slides)

limit of f x

function f x

values of f x

limit lim f x

cy c x

x0 x v63

nyu section

basic limits likelim

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