Lesson 6: Limits Involving ∞ (Section 41 slides)

Section 1.6Limits involving Infinity

V63.0121.041, Calculus I

New York University

September 22, 2010

Announcements

I Quiz 1 is next week in recitation. Covers Sections 1.1–1.4

. . . . . .

. . . . . .

Announcements

I Quiz 1 is next week inrecitation. Covers Sections1.1–1.4

V63.0121.041, Calculus I (NYU) Section 1.6 Limits involving Infinity September 22, 2010 2 / 37

. . . . . .

Objectives

I “Intuit” limits involvinginfinity by eyeballing theexpression.

I Show limits involvinginfinity by algebraicmanipulation andconceptual argument.


. . . . . .

Recall the definition of 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.


. . . . . .

Recall the unboundedness problem

Recall why limx→0+

1xdoesn’t exist.

. .x

.y

..L?

No matter how thin we draw the strip to the right of x = 0, we cannot“capture” the graph inside the box.


. . . . . .



1xdoesn’t exist.

. .x

.y

..L?



. . . . . .



1xdoesn’t exist.

. .x

.y

..L?



. . . . . .



1xdoesn’t exist.

. .x

.y

..L?



. . . . . .

Outline

Infinite LimitsVertical AsymptotesInfinite Limits we KnowLimit “Laws” with Infinite LimitsIndeterminate Limit forms

Limits at ∞Algebraic rates of growthRationalizing to get a limit


. . . . . .

Infinite Limits

DefinitionThe notation

limx→a

f(x) = ∞

means that values of f(x) canbe made arbitrarily large (aslarge as we please) by taking xsufficiently close to a but notequal to a.

I “Large” takes the place of“close to L”.

. .x

.y


. . . . . .

Infinite Limits


limx→a

f(x) = ∞



. .x

.y


. . . . . .

Infinite Limits


limx→a

f(x) = ∞



. .x

.y


. . . . . .

Infinite Limits


limx→a

f(x) = ∞



. .x

.y


. . . . . .

Infinite Limits


limx→a

f(x) = ∞



. .x

.y


. . . . . .

Infinite Limits


limx→a

f(x) = ∞



. .x

.y


. . . . . .

Infinite Limits


limx→a

f(x) = ∞



. .x

.y


. . . . . .

Infinite Limits


limx→a

f(x) = ∞



. .x

.y


. . . . . .

Negative Infinity


limx→a

f(x) = −∞

means that the values of f(x) can be made arbitrarily large negative (aslarge as we please) by taking x sufficiently close to a but not equal to a.

I We call a number large or small based on its absolute value. So−1,000,000 is a large (negative) number.


. . . . . .

Negative Infinity


limx→a

f(x) = −∞

means that the values of f(x) can be made arbitrarily large negative (aslarge as we please) by taking x sufficiently close to a but not equal to a.

I We call a number large or small based on its absolute value. So−1,000,000 is a large (negative) number.


. . . . . .

Vertical Asymptotes

DefinitionThe line x = a is called a vertical asymptote of the curve y = f(x) if atleast one of the following is true:

I limx→a

f(x) = ∞

I limx→a+

f(x) = ∞

I limx→a−

f(x) = ∞

I limx→a

f(x) = −∞

I limx→a+

f(x) = −∞

I limx→a−

f(x) = −∞


. . . . . .

Infinite Limits we Know

I limx→0+

1x= ∞

I limx→0−

1x= −∞

I limx→0

1x2

= ∞

. .x

.y

.

.

.

.

.

.

.

.

.

.

.

.


. . . . . .


I limx→0+

1x= ∞

I limx→0−

1x= −∞

I limx→0

1x2

= ∞

. .x

.y

.

.

.

.

.

.

.

.

.

.

.

.


. . . . . .


I limx→0+

1x= ∞

I limx→0−

1x= −∞

I limx→0

1x2

= ∞

. .x

.y

.

.

.

.

.

.

.

.

.

.

.

.


. . . . . .

Finding limits at trouble spots

Example

Let

f(x) =x2 + 2

x2 − 3x+ 2Find lim

x→a−f(x) and lim

x→a+f(x) for each a at which f is not continuous.

SolutionThe denominator factors as (x− 1)(x− 2). We can record the signs ofthe factors on the number line.


. . . . . .

Finding limits at trouble spots

Example

Let

f(x) =x2 + 2

x2 − 3x+ 2Find lim

x→a−f(x) and lim

x→a+f(x) for each a at which f is not continuous.

SolutionThe denominator factors as (x− 1)(x− 2). We can record the signs ofthe factors on the number line.


. . . . . .

Use the number line

. .(x− 1)

.− ..1.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+

.+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞

.−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞

.− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .−

.−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞

.+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞

.+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

Solim

x→1−f(x) = +∞

limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

Solim

x→1−f(x) = +∞

limx→2−

f(x) = −∞

limx→1+

f(x) = −∞

limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

Solim

x→1−f(x) = +∞ lim

x→2−f(x) = −∞

limx→1+

f(x) = −∞

limx→2+

f(x) = +∞


. . . . . .

Use the number line

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

Solim

x→1−f(x) = +∞ lim

x→2−f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞


. . . . . .

In English, now

To explain the limit, you can say:“As x → 1−, the numerator approaches 3, and the denominatorapproaches 0 while remaining positive. So the limit is +∞.”


. . . . . .

The graph so far

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. .x

.y

..−1

..1

..2

..3


. . . . . .

The graph so far

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. .x

.y

..−1

..1

..2

..3


. . . . . .

The graph so far

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. .x

.y

..−1

..1

..2

..3


. . . . . .

The graph so far

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. .x

.y

..−1

..1

..2

..3


. . . . . .

The graph so far

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. .x

.y

..−1

..1

..2

..3


. . . . . .

Limit Laws (?) with infinite limits

Fact

I If limx→a

f(x) = ∞ and limx→a

g(x) = ∞,

then limx→a

(f(x) + g(x)) = ∞.

.

.∞+∞ = ∞

I If limx→a

f(x) = −∞ andlimx→a

g(x) = −∞, then

limx→a

(f(x) + g(x)) = −∞.

.

.−∞+ (−∞) = −∞


. . . . . .

Rules of Thumb with infinite limits

Fact

I If limx→a


g(x) = ∞,

then limx→a

(f(x) + g(x)) = ∞..

.∞+∞ = ∞

I If limx→a


g(x) = −∞, then

limx→a

(f(x) + g(x)) = −∞.

.

.−∞+ (−∞) = −∞


. . . . . .


Fact

I If limx→a


g(x) = ∞,

then limx→a

(f(x) + g(x)) = ∞..

.∞+∞ = ∞

I If limx→a


g(x) = −∞, then

limx→a

(f(x) + g(x)) = −∞..

.−∞+ (−∞) = −∞


. . . . . .


Fact

I If limx→a

f(x) = L and limx→a

g(x) = ±∞,

.

.L+∞ = ∞L−∞ = −∞

then limx→a

(f(x) + g(x)) = ±∞.


. . . . . .


Fact

I If limx→a

f(x) = L and limx→a

g(x) = ±∞,.

.L+∞ = ∞L−∞ = −∞

then limx→a

(f(x) + g(x)) = ±∞.


. . . . . .

Rules of Thumb with infinite limitsKids, don't try this at home!

Fact

I The product of a finite limit and an infinite limit is infinite if the finitelimit is not 0.

.

.L · ∞ =

{∞ if L > 0−∞ if L < 0.

.

.L · (−∞) =

{−∞ if L > 0∞ if L < 0.


. . . . . .


Fact

I The product of a finite limit and an infinite limit is infinite if the finitelimit is not 0. .

.L · ∞ =

{∞ if L > 0−∞ if L < 0.

.

.L · (−∞) =

{−∞ if L > 0∞ if L < 0.


. . . . . .


Fact

I The product of a finite limit and an infinite limit is infinite if the finitelimit is not 0. .

.L · ∞ =

{∞ if L > 0−∞ if L < 0.

.

.L · (−∞) =

{−∞ if L > 0∞ if L < 0.


. . . . . .

Multiplying infinite limitsKids, don't try this at home!

Fact

I The product of two infinite limits is infinite.

.

.∞ ·∞ = ∞

∞ · (−∞) = −∞(−∞) · (−∞) = ∞


. . . . . .

Multiplying infinite limitsKids, don't try this at home!

Fact

I The product of two infinite limits is infinite. .

.∞ ·∞ = ∞

∞ · (−∞) = −∞(−∞) · (−∞) = ∞


. . . . . .

Dividing by InfinityKids, don't try this at home!

Fact

I The quotient of a finite limit by an infinite limit is zero.

.

.L∞

= 0


. . . . . .

Dividing by InfinityKids, don't try this at home!

Fact

I The quotient of a finite limit by an infinite limit is zero..

.L∞

= 0


. . . . . .

Dividing by zero is still not allowed

..10= ∞

There are examples of such limit forms where the limit is ∞, −∞,undecided between the two, or truly neither.


. . . . . .

Indeterminate Limit forms

Limits of the formL0are indeterminate. There is no rule for evaluating

such a form; the limit must be examined more closely. Consider these:

limx→0

1x2

= ∞ limx→0

−1x2

= −∞

limx→0+

1x= ∞ lim

x→0−1x= −∞

Worst, limx→0

1x sin(1/x)

is of the formL0, but the limit does not exist, even

in the left- or right-hand sense. There are infinitely many verticalasymptotes arbitrarily close to 0!


. . . . . .

Indeterminate Limit forms

Limits of the form 0 · ∞ and ∞−∞ are also indeterminate.

Example

I The limit limx→0+

sin x · 1xis of the form 0 · ∞, but the answer is 1.


sin2 x · 1xis of the form 0 · ∞, but the answer is 0.


sin x · 1x2

is of the form 0 · ∞, but the answer is ∞.

Limits of indeterminate forms may or may not “exist.” It will depend onthe context.


. . . . . .

Indeterminate forms are like Tug Of War

Which side wins depends on which side is stronger.


. . . . . .

Outline

Infinite LimitsVertical AsymptotesInfinite Limits we KnowLimit “Laws” with Infinite LimitsIndeterminate Limit forms

Limits at ∞Algebraic rates of growthRationalizing to get a limit


. . . . . .

DefinitionLet f be a function defined on some interval (a,∞). Then

limx→∞

f(x) = L

means that the values of f(x) can be made as close to L as we like, bytaking x sufficiently large.

DefinitionThe line y = L is a called a horizontal asymptote of the curve y = f(x)if either

limx→∞

f(x) = L or limx→−∞

f(x) = L.

y = L is a horizontal line!


. . . . . .


limx→∞

f(x) = L



limx→∞


f(x) = L.



. . . . . .


limx→∞

f(x) = L



limx→∞


f(x) = L.



. . . . . .

Basic limits at infinity

TheoremLet n be a positive integer. Then

I limx→∞

1xn

= 0

I limx→−∞

1xn

= 0


. . . . . .

Limit laws at infinity

FactAny limit law that concerns finite limits at a finite point a is still true ifthe finite point is replaced by infinity.That is, if lim

x→∞f(x) = L and lim

x→∞g(x) = M, then

I limx→∞

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

I limx→∞

(f(x)− g(x)) = L−M

I limx→∞

cf(x) = c · L (for any constant c)

I limx→∞

f(x) · g(x) = L ·M

I limx→∞

f(x)g(x)

=LM

(if M ̸= 0)


. . . . . .

Using the limit laws to compute limits at ∞

Example

Find limx→∞

xx2 + 1

AnswerThe limit is 0.

. .x

.y

Notice that the graph does cross the asymptote, which contradicts oneof the commonly held beliefs of what an asymptote is.


. . . . . .


Example

Find limx→∞

xx2 + 1


. .x

.y



. . . . . .

Solution

SolutionFactor out the largest power of x from the numerator and denominator.We have

xx2 + 1

=x(1)

x2(1+ 1/x2)=

1x· 11+ 1/x2

limx→∞

xx2 + 1

= limx→∞

1x

11+ 1/x2

= limx→∞

1x· limx→∞

11+ 1/x2

= 0 · 11+ 0

= 0.

RemarkHad the higher power been in the numerator, the limit would have been∞.


. . . . . .


Example

Find limx→∞

xx2 + 1


. .x

.y



. . . . . .

Solution


xx2 + 1

=x(1)

x2(1+ 1/x2)=

1x· 11+ 1/x2

limx→∞

xx2 + 1

= limx→∞

1x

11+ 1/x2

= limx→∞

1x· limx→∞

11+ 1/x2

= 0 · 11+ 0

= 0.

RemarkHad the higher power been in the numerator, the limit would have been∞.


. . . . . .

Another Example

Example

Find

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

if it exists.A does not existB 1/2

C 0D ∞


. . . . . .

Another Example

Example

Find

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

if it exists.A does not existB 1/2

C 0D ∞


. . . . . .

Solution


2x3 + 3x+ 14x3 + 5x2 + 7

=x3(2+ 3/x2 + 1/x3)

x3(4+ 5/x + 7/x3)

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

= limx→∞

2+ 3/x2 + 1/x3

4+ 5/x + 7/x3

=2+ 0+ 04+ 0+ 0

=12

Upshot

When finding limits of algebraic expressions at infinity, look at thehighest degree terms.


. . . . . .

Solution


2x3 + 3x+ 14x3 + 5x2 + 7

=x3(2+ 3/x2 + 1/x3)

x3(4+ 5/x + 7/x3)

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

= limx→∞

2+ 3/x2 + 1/x3

4+ 5/x + 7/x3

=2+ 0+ 04+ 0+ 0

=12

Upshot

When finding limits of algebraic expressions at infinity, look at thehighest degree terms.


. . . . . .

Still Another Example

Example

Find

limx→∞

√3x4 + 7x2 + 3

.

.√

3x4 + 7 ∼√

3x4 =√3x2

AnswerThe limit is

√3.


. . . . . .

Still Another Example

Example

Find

limx→∞

√3x4 + 7x2 + 3

.

.√

3x4 + 7 ∼√

3x4 =√3x2

AnswerThe limit is

√3.


. . . . . .

Solution

Solution

limx→∞

√3x4 + 7x2 + 3

= limx→∞

√x4(3+ 7/x4)

x2(1+ 3/x2)

= limx→∞

x2√

(3+ 7/x4)

x2(1+ 3/x2)

= limx→∞

√(3+ 7/x4)

1+ 3/x2

=

√3+ 01+ 0

=√3.


. . . . . .

Rationalizing to get a limit

Example

Compute limx→∞

(√4x2 + 17− 2x

).

SolutionThis limit is of the form ∞−∞, which we cannot use. So werationalize the numerator (the denominator is 1) to get an expressionthat we can use the limit laws on.

limx→∞

(√4x2 + 17− 2x

)= lim

x→∞

(√4x2 + 17− 2x

)·√4x2 + 17+ 2x√4x2 + 17+ 2x

= limx→∞

(4x2 + 17)− 4x2√4x2 + 17+ 2x

= limx→∞

17√4x2 + 17+ 2x

= 0


. . . . . .

Rationalizing to get a limit

Example

Compute limx→∞

(√4x2 + 17− 2x

).

SolutionThis limit is of the form ∞−∞, which we cannot use. So werationalize the numerator (the denominator is 1) to get an expressionthat we can use the limit laws on.

limx→∞

(√4x2 + 17− 2x

)= lim

x→∞

(√4x2 + 17− 2x

)·√4x2 + 17+ 2x√4x2 + 17+ 2x

= limx→∞

(4x2 + 17)− 4x2√4x2 + 17+ 2x

= limx→∞

17√4x2 + 17+ 2x

= 0


. . . . . .

Kick it up a notch

Example

Compute limx→∞

(√4x2 + 17x− 2x

).

SolutionSame trick, different answer:

limx→∞

(√4x2 + 17x− 2x

)= lim

x→∞

(√4x2 + 17x− 2x

)·√4x2 + 17+ 2x√4x2 + 17x+ 2x

= limx→∞

(4x2 + 17x)− 4x2√4x2 + 17x+ 2x

= limx→∞

17x√4x2 + 17x+ 2x

= limx→∞

17√4+ 17/x+ 2

=174


. . . . . .

Kick it up a notch

Example

Compute limx→∞

(√4x2 + 17x− 2x

).

SolutionSame trick, different answer:

limx→∞

(√4x2 + 17x− 2x

)= lim

x→∞

(√4x2 + 17x− 2x

)·√4x2 + 17+ 2x√4x2 + 17x+ 2x

= limx→∞

(4x2 + 17x)− 4x2√4x2 + 17x+ 2x

= limx→∞

17x√4x2 + 17x+ 2x

= limx→∞

17√4+ 17/x+ 2

=174


. . . . . .

Summary

I Infinity is a more complicated concept than a single number.There are rules of thumb, but there are also exceptions.

I Take a two-pronged approach to limits involving infinity:I Look at the expression to guess the limit.I Use limit rules and algebra to verify it.


Documents

Lesson 6: Limits Involving ∞ (Section 41 slides)