Lesson 3: Limit Laws

Section 1.4Calculating Limits

V63.0121.002.2010Su, Calculus I

New York University

May 18, 2010

Announcements

I WebAssign Class Key: nyu 0127 7953I Office Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here)I Quiz 1 Thursday on 1.1–1.4

. . . . . .

. . . . . .

Announcements

I WebAssign Class Key: nyu0127 7953

I Office Hours: MR5:00–5:45, TW 7:50–8:30,CIWW 102 (here)

I Quiz 1 Thursday on1.1–1.4

V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 2 / 37

. . . . . .

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

Basic Limits

Limit LawsThe direct substitution property

Limits with AlgebraTwo more limit theorems

Two important trigonometric 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.


. . . . . .

Really basic limits


x→ax = a

(ii) limx→a

c = c



. . . . . .

ET game for f(x) = x

. .x

.y

..a

..a

I Setting error equal to tolerance works!


. . . . . .


. .x

.y

..a

..a



. . . . . .


. .x

.y

..a

..a



. . . . . .


. .x

.y

..a

..a



. . . . . .


. .x

.y

..a

..a



. . . . . .


. .x

.y

..a

..a



. . . . . .


. .x

.y

..a

..a



. . . . . .

ET game for f(x) = c

.

.x

.y

..a

..c

I any tolerance works!


. . . . . .


. .x

.y

..a

..c



. . . . . .


. .x

.y

..a

..c



. . . . . .


. .x

.y

..a

..c



. . . . . .


. .x

.y

..a

..c



. . . . . .


. .x

.y

..a

..c



. . . . . .


. .x

.y

..a

..c



. . . . . .

Really basic limits


x→ax = a

(ii) limx→a

c = c



. . . . . .

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



. . . . . .


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 Player 2 gives us an error of 1 (for instance), Player 1 can

use the fact that limx→a

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

I Player 1 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, it can.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)


. . . . . .


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)


. . . . . .


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

)

= 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.


. . . . . .


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.


. . . . . .

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.


. . . . . .




x→af(x) = L.

Example

Find limx→−1

x2 + 2x+ 1x+ 1

, if it exists.

Solution

Sincex2 + 2x+ 1


limx→−1


x2 + 2x+ 1x+ 1

= 0.


. . . . . .




x→af(x) = L.

Example

Find limx→−1

x2 + 2x+ 1x+ 1

, if it exists.

Solution

Sincex2 + 2x+ 1


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.


. . . . . .

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.


. . . . . .


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.V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 24 / 37

. . . . . .

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



. . . . . .

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



. . . . . .

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



. . . . . .

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



. . . . . .

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.V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 26 / 37

. . . . . .

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.


. . . . . .



Example

Show that limx→0

x2 sin(πx

)= 0.


−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2



. . . . . .



Example

Show that limx→0

x2 sin(πx

)= 0.


−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2



. . . . . .

Illustration of the Squeeze Theorem

. .x

.y .h(x) = x2

.f(x) = −x2

.g(x) = x2 sin(πx

)


. . . . . .


. .x

.y .h(x) = x2

.f(x) = −x2

.g(x) = x2 sin(πx

)


. . . . . .


. .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.

. .θ.sin 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 θθ

≥ cos θ



. . . . . .


Proof.

. .θ.sin 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 θθ

≥ cos θ



. . . . . .


Proof.

. .θ.sin 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 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


. . . . . .

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 · 12

= 2 · lim2θ→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


. . . . . .

Solutions


limθ→0

tan θθ

= limθ→0

sin θθ cos θ

= limθ→0

sin θθ

· limθ→0

1cos θ

= 1 · 11= 1.


limθ→0

sin 2θθ

= lim2θ→0

sin 2θ2θ · 12

= 2 · lim2θ→0

sin 2θ2θ

= 2 · 1 = 2


limθ→0

sin 2θθ

= limθ→0

2 sin θ cos θθ

= 2 · limθ→0

sin θθ

· limθ→0

cos θ = 2 ·1 ·1 = 2


. . . . . .

Solutions


limθ→0

tan θθ

= limθ→0

sin θθ cos θ

= limθ→0

sin θθ

· limθ→0

1cos θ

= 1 · 11= 1.


limθ→0

sin 2θθ

= lim2θ→0

sin 2θ2θ · 12

= 2 · lim2θ→0

sin 2θ2θ

= 2 · 1 = 2


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 Limits laws say limits playwell with the rules ofarithmetic

I When limit laws do notwork we can bealgebraically creative

I When algebra does notwork we can trySqueezing.


Lesson 3: Limit Laws

Education

lim fx gx

lim gx

c fx c

arithmetic factsuppose

axaii lim c

lim cfx

lim fxgx

nyu section