Lesson 5: Limits Involving Infinity

Section 1.6Limits involving Infinity

V63.0121.002.2010Su, Calculus I

New York University

May 20, 2010

Announcements

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

I Quiz 1 Thursday on 1.1–1.4

V63.0121.002.2010Su, Calculus I (NYU) Section 1.6 Limits involving Infinity May 20, 2010 2 / 37

Objectives

I “Intuit” limits involvinginfinity by eyeballing theexpression.

I Show limits involving infinityby algebraic manipulationand conceptual argument.


Recall the definition of 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.


Recall the unboundedness problem

Recall why limx→0+

1

xdoesn’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.




1

xdoesn’t exist.

x

y

L?





1

xdoesn’t exist.

x

y

L?





1

xdoesn’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

Definition

The notation

limx→a

f (x) =∞

means that values of f (x) can bemade arbitrarily large (as large aswe please) by taking x sufficientlyclose to a but not equal to a.

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

x

y


Infinite Limits

Definition

The notation

limx→a

f (x) =∞



x

y


Infinite Limits

Definition

The notation

limx→a

f (x) =∞



x

y


Infinite Limits

Definition

The notation

limx→a

f (x) =∞



x

y


Infinite Limits

Definition

The notation

limx→a

f (x) =∞



x

y


Infinite Limits

Definition

The notation

limx→a

f (x) =∞



x

y


Infinite Limits

Definition

The notation

limx→a

f (x) =∞



x

y


Infinite Limits

Definition

The notation

limx→a

f (x) =∞



x

y


Negative Infinity

Definition

The notationlimx→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

Definition

The notationlimx→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

Definition

The 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 lim

x→a+f (x) =∞

I limx→a−

f (x) =∞

I limx→a

f (x) = −∞I lim

x→a+f (x) = −∞

I limx→a−

f (x) = −∞


Infinite Limits we Know

I limx→0+

1

x=∞

I limx→0−

1

x= −∞

I limx→0

1

x2=∞

x

y



I limx→0+

1

x=∞

I limx→0−

1

x= −∞

I limx→0

1

x2=∞

x

y



I limx→0+

1

x=∞

I limx→0−

1

x= −∞

I limx→0

1

x2=∞

x

y


Finding limits at trouble spots

Example

Let

f (x) =x2 + 2

x2 − 3x + 2

Find limx→a−

f (x) and limx→a+

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

Solution

The denominator factors as (x − 1)(x − 2). We can record the signs of thefactors on the number line.


Finding limits at trouble spots

Example

Let

f (x) =x2 + 2

x2 − 3x + 2

Find limx→a−

f (x) and limx→a+

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

Solution

The denominator factors as (x − 1)(x − 2). We can record the signs of thefactors 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

+ +∞

−∞ − −∞ +∞ +

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) = +∞

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) = +∞


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

x

y

−1 1 2 3


The graph so far

x

y

−1 1 2 3


The graph so far

x

y

−1 1 2 3


The graph so far

x

y

−1 1 2 3


The graph so far

x

y

−1 1 2 3


Limit Laws (?) with infinite limits

I If limx→a

f (x) =∞ and limx→a

g(x) =∞, then limx→a

(f (x) + g(x)) =∞.

That is,

∞+∞ =∞

I If limx→a

f (x) = −∞ and limx→a

g(x) = −∞, then

limx→a

(f (x) + g(x)) = −∞. That is,

−∞−∞ = −∞


Rules of Thumb with infinite limits

I If limx→a

f (x) =∞ and limx→a

g(x) =∞, then limx→a

(f (x) + g(x)) =∞.

That is,

∞+∞ =∞

I If limx→a

f (x) = −∞ and limx→a

g(x) = −∞, then

limx→a

(f (x) + g(x)) = −∞. That is,

−∞−∞ = −∞


Rules of Thumb with infinite limits

I If limx→a

f (x) = L and limx→a

g(x) = ±∞, then limx→a

(f (x) + g(x)) = ±∞.

That is,

L +∞ =∞L−∞ = −∞


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

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!

I The product of two infinite limits is infinite.

∞ ·∞ =∞∞ · (−∞) = −∞

(−∞) · (−∞) =∞


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

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

L

∞= 0


Dividing by zero is still not allowed

1

0=∞

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


Indeterminate Limit forms

Limits of the formL

0are indeterminate. There is no rule for evaluating

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

limx→0

1

x2=∞ lim

x→0

−1

x2= −∞

limx→0+

1

x=∞ lim

x→0−

1

x= −∞

Worst, limx→0

1

x sin(1/x)is of the form

L

0, 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 · 1

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


sin2 x · 1

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


sin x · 1

x2is 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


Limits at infinity

Definition

Let 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.

Definition

The 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!


Limits at infinity

Definition


limx→∞

f (x) = L


Definition


limx→∞


f (x) = L.



Limits at infinity

Definition


limx→∞

f (x) = L


Definition


limx→∞


f (x) = L.



Basic limits at infinity

Theorem

Let n be a positive integer. Then

I limx→∞

1

xn= 0

I limx→−∞

1

xn= 0


Using the limit laws to compute limits at ∞

Example

Find

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7

if it exists.

A does not exist

B 1/2

C 0

D ∞


Using the limit laws to compute limits at ∞

Example

Find

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7

if it exists.

A does not exist

B 1/2

C 0

D ∞


Solution

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

2x3 + 3x + 1

4x3 + 5x2 + 7=

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

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

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7= lim

x→∞

2 + 3/x2 + 1/x3

4 + 5/x + 7/x3

=2 + 0 + 0

4 + 0 + 0=

1

2

Upshot

When finding limits of algebraic expressions at infinity, look at the highestdegree terms.


Solution

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

2x3 + 3x + 1

4x3 + 5x2 + 7=

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

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

limx→∞

2x3 + 3x + 1

4x3 + 5x2 + 7= lim

x→∞

2 + 3/x2 + 1/x3

4 + 5/x + 7/x3

=2 + 0 + 0

4 + 0 + 0=

1

2

Upshot

When finding limits of algebraic expressions at infinity, look at the highestdegree terms.


Another Example

Example

Find limx→∞

x

x2 + 1

Answer

The limit is 0.

x

y

Notice that the graph does cross the asymptote, which contradicts one ofthe heuristic definitions of asymptote.


Another Example

Example

Find limx→∞

x

x2 + 1

Answer

The limit is 0.

x

y



Solution

Again, factor out the largest power of x from the numerator anddenominator. We have

x

x2 + 1=

x(1)

x2(1 + 1/x2)=

1

x· 1

1 + 1/x2

limx→∞

x

x2 + 1= lim

x→∞

1

x

1

1 + 1/x2= lim

x→∞

1

x· limx→∞

1

1 + 1/x2

= 0 · 1

1 + 0= 0.

Remark

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


Another Example

Example

Find limx→∞

x

x2 + 1

Answer

The limit is 0.

x

y



Another Example

Example

Find limx→∞

x

x2 + 1

Answer

The limit is 0.

x

y



Solution

Again, factor out the largest power of x from the numerator anddenominator. We have

x

x2 + 1=

x(1)

x2(1 + 1/x2)=

1

x· 1

1 + 1/x2

limx→∞

x

x2 + 1= lim

x→∞

1

x

1

1 + 1/x2= lim

x→∞

1

x· limx→∞

1

1 + 1/x2

= 0 · 1

1 + 0= 0.

Remark

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


Another Example

Example

Find

limx→∞

√3x4 + 7

x2 + 3

√3x4 + 7 ∼

√3x4 =

√3x2

Answer

The limit is√

3.


Another Example

Example

Find

limx→∞

√3x4 + 7

x2 + 3

√3x4 + 7 ∼

√3x4 =

√3x2

Answer

The limit is√

3.


Solution

limx→∞

√3x4 + 7

x2 + 3= lim

x→∞

√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 + 0

1 + 0=√

3.


Rationalizing to get a limit

Example

Compute limx→∞

(√4x2 + 17− 2x

).

Solution

This limit is of the form ∞−∞, which we cannot use. So we rationalizethe numerator (the denominator is 1) to get an expression that we can usethe 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

).

Solution

This limit is of the form ∞−∞, which we cannot use. So we rationalizethe numerator (the denominator is 1) to get an expression that we can usethe 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

).

Solution

Same trick, different answer:

limx→∞

(√4x2 + 17x − 2x

)= lim

x→∞

(√4x2 + 17x − 2x

)·√

4x2 + 17x + 2x√4x2 + 17x + 2x

= limx→∞

(4x2 + 17x)− 4x2

√4x2 + 17x + 2x

= limx→∞

17x√4x2 + 17x + 2x

= limx→∞

17√4 + 17/x + 2

=17

4


Kick it up a notch

Example

Compute limx→∞

(√4x2 + 17x − 2x

).

Solution

Same trick, different answer:

limx→∞

(√4x2 + 17x − 2x

)= lim

x→∞

(√4x2 + 17x − 2x

)·√

4x2 + 17x + 2x√4x2 + 17x + 2x

= limx→∞

(4x2 + 17x)− 4x2

√4x2 + 17x + 2x

= limx→∞

17x√4x2 + 17x + 2x

= limx→∞

17√4 + 17/x + 2

=17

4


Summary

I Infinity is a morecomplicated concept than asingle number. There arerules of thumb, but there arealso exceptions.

I Take a two-prongedapproach to limits involvinginfinity:

I Look at the expression toguess the limit.

I Use limit rules andalgebra to verify it.


Education

Lesson 5: Limits Involving Infinity