Lesson 6: Limits Involving Infinity (handout)

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Sec on 1.5Limits Involving Infinity

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

February 9, 2011

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Announcements

I Get-to-know-you extracredit due FridayFebruary 11

I Quiz 1 is next week inrecita on. CoversSec ons 1.1–1.4

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Objectives

I “Intuit” limits involving infinity byeyeballing the expression.

I Show limits involving infinity byalgebraic manipula on and conceptualargument.

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. Sec on 1.5: Limits Involving Infinity. V63.0121.001: Calculus I . February 9, 2011

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Recall the definition of limitDefini onWe 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 Las we like) by taking x to be sufficiently close to a (on either side ofa) but not equal to a.

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The unboundedness problem

Recall why limx→0+

1xdoesn’t

exist.No ma er how thin we drawthe strip to the right of x = 0,we cannot “capture” the graphinside the box.

.. x.

y

..

L?

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Infinite LimitsDefini onThe nota on

limx→a

f(x) = ∞

means that values of f(x) can bemade arbitrarily large (as large as weplease) by taking x sufficiently closeto a but not equal to a. .. x.

y

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

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Negative InfinityDefini onThe nota on

limx→a

f(x) = −∞

means that the values of f(x) can be made arbitrarily large nega ve(as large as we please) by taking x sufficiently close to a but notequal to a.

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

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Vertical Asymptotes

Defini onThe line x = a is called a ver cal asymptote of the curve y = f(x) ifat least 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) = −∞

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Infinite Limits we Know

I limx→0+

1x= ∞

I limx→0−

1x= −∞

I limx→0

1x2

= ∞

.. x.

y

............

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Finding limits at trouble spotsExample

Letf(x) =

x2 + 2x2 − 3x+ 2

Find limx→a−

f(x) and limx→a+

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

Solu onThe denominator factors as (x− 1)(x− 2). We can record the signsof the factors on the number line.

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Use the number line

.. (x− 1).−. small

. small

..1

. 0. +.

(x− 2)

.−..

2

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0

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small.

small.

+

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(x2 + 2)

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+

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f(x)

..

1

..

2

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+

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+∞

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−∞

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−

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−∞

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+∞

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+

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

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In English, now

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

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The graph so farlimx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

.. x.

y

..−1

..1

..2

..3

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Rules of Thumb with infinite limitsFactThe sum of two posi ve or two nega ve infinite limits is infinite.

I If limx→a

f(x) = ∞ and limx→a

g(x) = ∞, then limx→a

(f(x) + g(x)) = ∞.

..

∞+∞ = ∞

I If limx→a

f(x) = −∞ and limx→a

g(x) = −∞, thenlimx→a

(f(x) + g(x)) = −∞.

..

−∞+ (−∞) = −∞

RemarkWe don’t say anything here about limits of the form∞−∞.

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Rules of Thumb with infinite limits

FactThe sum of a finite limit and an infinite limit is infinite.

I If limx→a

f(x) = L and limx→a

g(x) = ±∞,

..

L+∞ = ∞L−∞ = −∞

thenlimx→a

(f(x) + g(x)) = ±∞.

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Rules of Thumb with infinite limitsFactThe product of a finite limit and an infinite limit is infinite if the finitelimit is not 0.

...

L · ∞ =

{∞ if L > 0−∞ if L < 0.

I If limx→a

f(x) = L, limx→a

g(x) = ∞, and L > 0, thenlimx→a

f(x) · g(x) = ∞.

I If limx→a

f(x) = L, limx→a

g(x) = ∞, and L < 0, thenlimx→a

f(x) · g(x) = −∞.

I If limx→a

f(x) = L, limx→a

g(x) = −∞, and L > 0, thenlimx→a

f(x) · g(x) = −∞.

I If limx→a

f(x) = L, limx→a

g(x) = −∞, and L < 0, thenlimx→a

f(x) · g(x) = ∞.

..

L · (−∞) =

{−∞ if L > 0∞ if L < 0.

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Multiplying infinite limits

FactThe product of two infinite limits is infinite.

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∞ ·∞ = ∞∞ · (−∞) = −∞

(−∞) · (−∞) = ∞

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) = −∞.

I If limx→a

f(x) = −∞ and limx→a

g(x) = −∞, then limx→a

f(x) · g(x) = ∞.

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Dividing by Infinity

FactThe quo ent of a finite limit by an infinite limit is zero.

I If limx→a

f(x) = L and limx→a

g(x) = ±∞, then limx→a

f(x)g(x)

= 0.

..

L∞

= 0

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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.

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Indeterminate Limit formsLimits of the form

L0are indeterminate. There is no rule for

evalua ng 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 le - or right-hand sense. There are infinitely manyver cal asymptotes arbitrarily close to 0!

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Indeterminate Limit formsLimits 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 dependon the context.

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Indeterminate forms are like Tug Of War

Which side wins depends on which side is stronger.

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

Limits at∞Algebraic rates of growthRa onalizing to get a limit

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Limits at Infinity

Defini onLet f be a func on 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.

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Horizontal Asymptotes

Defini onThe line y = L is a called a horizontal asymptote of the curvey = f(x) if either

limx→∞

f(x) = L or limx→−∞

f(x) = L.

y = L is a horizontal line!

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Basic limits at infinity

TheoremLet n be a posi ve integer. Then

I limx→∞

1xn

= 0

I limx→−∞

1xn

= 0

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Limit laws at infinityFactAny limit law that concerns finite limits at a finite point a is s ll trueif the finite point is replaced by±∞.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 lim

x→∞f(x) · g(x) = L ·M

I limx→∞

f(x)g(x)

=LM

(if M ̸= 0), etc.

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Using the limit laws to computelimits at ∞

Example

Find limx→∞

xx2 + 1

AnswerThe limit is 0. No ce thatthe graph does cross theasymptote, whichcontradicts one of thecommonly held beliefs ofwhat an asymptote is.

.. x.

y

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SolutionSolu on

Factor out the largest power of x from the numerator anddenominator. 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 havebeen∞.

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Another ExampleExample

Findlimx→∞

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

if it exists.A does not existB 1/2

C 0D ∞

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SolutionSolu on

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

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.

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Still Another Example

Example

Find

limx→∞

√3x4 + 7x2 + 3

..

√3x4 + 7 ∼

√3x4 =

√3x2

AnswerThe limit is

√3.

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SolutionSolu on

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.

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Rationalizing to get a limit..Example

Compute limx→∞

(√4x2 + 17− 2x

).

Solu onThis limit is of the form∞−∞, which we cannot use. So we ra onalize thenumerator (the denominator is 1) to get an expression that we can use thelimit 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

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Kick it up a notchExample

Compute limx→∞

(√4x2 + 17x− 2x

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Solu on

Same 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

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Summary

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

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.

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Lesson 6: Limits Involving Infinity (handout)