Lesson 6: Limits Involving Infinity (slides)

Sec on 1.6
Limits Involving Inﬁnity
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

February 9, 2011

.

Announcements

Get-to-know-you extra
credit due Friday
February 11
Quiz 1 is next week in
recita on. Covers
Sec ons 1.1–1.4

Objectives

“Intuit” limits involving inﬁnity by
eyeballing the expression.
Show limits involving inﬁnity by
algebraic manipula on and conceptual
argument.

Recall the definition of limit
Defini on
We write
lim f(x) = L
x→a
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 we like) by taking x to be sufficiently close to a (on either side of
a) but not equal to a.

The unboundedness problem
y
1
Recall why lim+ doesn’t
x→0 x
exist.
No ma er how thin we draw
the strip to the right of x = 0, L?
we cannot “capture” the graph
inside the box.
. x

Outline
Infinite Limits
Ver cal Asymptotes
Infinite Limits we Know
Limit “Laws” with Infinite Limits
Indeterminate Limit forms

Limits at ∞
Algebraic rates of growth
Ra onalizing to get a limit

Infinite Limits
Defini on
y
The nota on

lim f(x) = ∞
x→a

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

Infinite Limits
Defini on
y
The nota on

lim f(x) = ∞
x→a

means that values of f(x) can be
made arbitrarily large (as large as we
please) by taking x sufficiently close
. x
to a but not equal to a.
“Large” takes the place of “close to L”.

Negative Infinity
Defini on
The nota on
lim f(x) = −∞
x→a

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 not
equal to a.

Negative Infinity
Defini on
The nota on
lim f(x) = −∞
x→a

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 not
equal to a.

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

Vertical Asymptotes
Deﬁni on
The line x = a is called a ver cal asymptote of the curve y = f(x) if
at least one of the following is true:
lim f(x) = ∞ lim f(x) = −∞
x→a x→a
lim f(x) = ∞ lim f(x) = −∞
x→a+ x→a+
lim f(x) = ∞ lim f(x) = −∞
x→a− x→a−

y

1
lim+ =∞
x→0 x

. x

y

1
lim+ =∞
x→0 x
1
lim− = −∞
x→0 x . x

y

1
lim+ =∞
x→0 x
1
lim− = −∞
x→0 x . x
1
lim 2 = ∞
x→0 x

Finding limits at trouble spots
Example
Let
x2 + 2
f(x) = 2
x − 3x + 2
Find lim− f(x) and lim+ f(x) for each a at which f is not con nuous.
x→a x→a

Finding limits at trouble spots
Example
Let
x2 + 2
f(x) = 2
x − 3x + 2
Find lim− f(x) and lim+ f(x) for each a at which f is not con nuous.
x→a x→a

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

Use the number line
. (x − 1)

Use the number line
−. 0 +
(x − 1)
1

Use the number line
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2

Use the number line
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x2 + 2)

Use the number line
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x2 + 2)
f(x)
1 2

Use the number line
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x2 + 2)
+
f(x)
1 2

Use the number line

small
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x2 + 2)
+ +∞
f(x)
1 2
lim f(x) = + ∞
x→1−

Use the number line

small
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x2 + 2)
+ +∞ −∞
f(x)
1 2
lim f(x) = + ∞
x→1−
lim f(x) = − ∞
x→1+

Use the number line
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x2 + 2)
+ +∞ −∞−
f(x)
1 2
lim f(x) = + ∞
x→1−
lim f(x) = − ∞
x→1+

Use the number line
−. 0 +
(x − 1)

small
1
− 0 +
(x − 2)
2
+
(x2 + 2)
+ +∞ −∞−−∞
f(x)
1 2
lim f(x) = + ∞ lim f(x) = − ∞
x→1− x→2−
lim f(x) = − ∞
x→1+

Use the number line
−. 0 +
(x − 1)

small
1
− 0 +
(x − 2)
2
+
(x2 + 2)
+ +∞ −∞−−∞ +∞
f(x)
1 2
lim f(x) = + ∞ lim f(x) = − ∞
x→1− x→2−
lim f(x) = − ∞ lim f(x) = + ∞
x→1+ x→2+

Use the number line
−. 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x2 + 2)
+ +∞ −∞−−∞ +∞ +
f(x)
1 2
lim f(x) = + ∞ lim f(x) = − ∞
x→1− x→2−
lim f(x) = − ∞ lim f(x) = + ∞
x→1+ x→2+

In English, now
To explain the limit, you can
say/write:

“As x → 1− , the numerator
approaches 3, and the
denominator approaches 0
while remaining posi ve. So
the limit is +∞.”

The graph so far
lim f(x) = + ∞ lim f(x) = − ∞
x→1− x→2−
lim f(x) = − ∞ lim f(x) = + ∞
x→1+ x→2+
y

. x
−1 1 2 3

Limit Laws (?) with infinite limits
Fact
The sum of two posi ve or two nega ve infinite limits is infinite.
If lim f(x) = ∞ and lim g(x) = ∞, then lim (f(x) + g(x)) = ∞.
x→a x→a x→a

If lim f(x) = −∞ and lim g(x) = −∞, then
x→a x→a
lim (f(x) + g(x)) = −∞.
x→a

Rules of Thumb with inﬁnite limits
Fact ∞+∞=∞
. x→a x→a x→a

x→a x→a
lim (f(x) + g(x)) = −∞.
x→a

Fact ∞+∞=∞
−∞ + (−∞) = −∞
. x→a x→a x→a

x→a x→a .
lim (f(x) + g(x)) = −∞.
x→a

Fact ∞+∞=∞
−∞ + (−∞) = −∞
. x→a x→a x→a

x→a x→a .
lim (f(x) + g(x)) = −∞.
x→a

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

Kids, don’t try this at home!

Fact
The sum of a finite limit and an infinite limit is infinite.
If lim f(x) = L and lim g(x) = ±∞, then
x→a x→a
lim (f(x) + g(x)) = ±∞.
x→a


Fact
The sum of a finite limit and an infinite limit is infinite.
.
If lim f(x) = L and lim g(x) = ±∞, then
x→a x→a
lim (f(x) + g(x)) = ±∞.
x→a L+∞=∞
L − ∞ = −∞

Fact
The product of a finite limit and an infinite limit is infinite if the finite
limit is not 0.
If lim f(x) = L, lim g(x) = ∞, and L > 0, then
x→a x→a
lim f(x) · g(x) = ∞.
x→a

Fact
limit is not 0.
If lim f(x) = L, lim g(x) = ∞, and L > 0, then
x→a x→a
lim f(x) · g(x) = ∞.
x→a
If lim f(x) = L, lim g(x) = ∞, and L < 0, then
x→a x→a
lim f(x) · g(x) = −∞.
x→a

Fact
limit is not 0. {
∞
If lim f(x) = L, lim g(x)· = ∞, and L > if Lthen
0, > 0
L ∞=
x→a x→a
−∞ if L < 0.
lim f(x) · g(x) = ∞.
x→a
If lim f(x) = L, lim g(x) = ∞, and L < 0, then
x→a x→a .
lim f(x) · g(x) = −∞.
x→a

Fact
limit is not 0.
If lim f(x) = L, lim g(x) = −∞, and L > 0, then
x→a x→a
lim f(x) · g(x) = −∞.
x→a

Fact
limit is not 0.
If lim f(x) = L, lim g(x) = −∞, and L > 0, then
x→a x→a
lim f(x) · g(x) = −∞.
x→a
If lim f(x) = L, lim g(x) = −∞, and L < 0, then
x→a x→a
lim f(x) · g(x) = ∞.
x→a

Fact
limit is not 0. {
If lim f(x) = L, lim g(x) = −∞, −∞ L > L > 0
L · (−∞) = and
if 0, then
x→a x→a
∞ if L < 0.
lim f(x) · g(x) = −∞.
x→a
If lim f(x) = L, lim g(x) = −∞, and L < 0, then
x→a x→a .
lim f(x) · g(x) = ∞.
x→a

Multiplying infinite limits

Fact
The product of two infinite limits is infinite.
If lim f(x) = ∞ and lim g(x) = ∞, then lim f(x) · g(x) = ∞.
x→a x→a x→a
If lim f(x) = ∞ and lim g(x) = −∞, then lim f(x) · g(x) = −∞.
x→a x→a x→a
If lim f(x) = −∞ and lim g(x) = −∞, then lim f(x) · g(x) = ∞.
x→a x→a x→a

Multiplying infinite limits
Kids, don’t try this at home! ∞ ·∞ · ∞ = ∞
(−∞) = −∞
(−∞) · (−∞) = ∞
Fact
The product of two infinite limits is infinite. .
If lim f(x) = ∞ and lim g(x) = ∞, then lim f(x) · g(x) = ∞.
x→a x→a x→a
If lim f(x) = ∞ and lim g(x) = −∞, then lim f(x) · g(x) = −∞.
x→a x→a x→a
If lim f(x) = −∞ and lim g(x) = −∞, then lim f(x) · g(x) = ∞.
x→a x→a x→a

Dividing by Infinity

Fact
The quo ent of a finite limit by an infinite limit is zero.
f(x)
If lim f(x) = L and lim g(x) = ±∞, then lim = 0.
x→a x→a x→a g(x)

Dividing by Infinity
L
=0
Fact ∞
The quo ent of a finite limit by an infinite limit is zero.
f(x) .
If lim f(x) = L and lim g(x) = ±∞, then lim = 0.
x→a x→a x→a g(x)

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
L
Limits of the form are indeterminate. There is no rule for
0
evalua ng such a form; the limit must be examined more closely.
Consider these:
1 −1
lim =∞ lim = −∞
x→0 x2 x→0 x2
1 1
lim+ = ∞ lim− = −∞
x→0 x x→0 x

1 L
Worst, lim is of the form , but the limit does not exist,
x→0 x sin(1/x) 0
even in the le - or right-hand sense. There are inﬁnitely many
ver cal asymptotes arbitrarily close to 0!

More Indeterminate Limit forms
Limits of the form 0 · ∞ and ∞ − ∞ are also indeterminate.
Example
1
The limit lim+ sin x · is of the form 0 · ∞, but the answer is 1.
x→0 x
1
The limit lim+ sin2 x · is of the form 0 · ∞, but the answer is 0.
x→0 x
1
The limit lim+ sin x · 2 is of the form 0 · ∞, but the answer is ∞.
x→0 x

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

Indeterminate forms are like Tug Of War

Which side wins depends on which side is stronger.

Limits at Infinity

Defini on
Let f be a func on defined on some interval (a, ∞). Then

lim f(x) = L
x→∞

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

Horizontal Asymptotes

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

lim f(x) = L or lim f(x) = L.
x→∞ x→−∞

Horizontal Asymptotes

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

lim f(x) = L or lim f(x) = L.
x→∞ x→−∞

y = L is a horizontal line!

Basic limits at inﬁnity

Theorem
Let n be a posi ve integer. Then
1
lim n = 0
x→∞ x
1
lim n = 0
x→−∞ x

Limit laws at infinity
Fact
Any limit law that concerns finite limits at a finite point a is s ll true
if the finite point is replaced by ±∞. That is, if lim f(x) = L and
x→∞
lim g(x) = M, then
x→∞

lim (f(x) + g(x)) = L + M
x→∞

Fact
x→∞
lim g(x) = M, then
x→∞

lim (f(x) + g(x)) = L + M
x→∞
lim (f(x) − g(x)) = L − M
x→∞

Fact
x→∞
lim g(x) = M, then
x→∞

lim (f(x) + g(x)) = L + M
x→∞
lim (f(x) − g(x)) = L − M
x→∞
lim cf(x) = c · L (for any
x→∞
constant c)

Fact
x→∞
lim g(x) = M, then
x→∞

lim (f(x) + g(x)) = L + M lim f(x) · g(x) = L · M
x→∞
x→∞
lim (f(x) − g(x)) = L − M
x→∞
lim cf(x) = c · L (for any
x→∞
constant c)

Fact
x→∞
lim g(x) = M, then
x→∞

x→∞
x→∞
lim (f(x) − g(x)) = L − M f(x) L
x→∞
lim = (if M ̸= 0)
lim cf(x) = c · L (for any x→∞ g(x) M
x→∞
constant c)

Fact
x→∞
lim g(x) = M, then
x→∞

x→∞
x→∞
lim (f(x) − g(x)) = L − M f(x) L
x→∞
lim = (if M ̸= 0)
lim cf(x) = c · L (for any x→∞ g(x) M
x→∞
constant c) etc.

Computing limits at ∞
With the limit laws
Example
x
Find lim if it exists.
x→∞ x2 + 1

Computing limits at ∞
With the limit laws
Example
x
x→∞ x2 + 1

Answer
The limit is 0. No ce that
the graph does cross the
y
asymptote, which
contradicts one of the .
commonly held beliefs of x
what an asymptote is.

Solution
Solu on
Factor out the largest power of x from the numerator and
denominator. We have
x x(1) 1 1
= 2 = · , so
x2 + 1 x (1 + 1/x2 ) x 1 + 1/x2
x 1 1 1 1 1
lim 2 = lim = lim · lim =0· = 0.
x→∞ x + 1 x→∞ x 1 + 1/x2 x→∞ x x→∞ 1 + 1/x2 1+0

Another Example

Example
x3 + 2x2 + 4
x→∞ 3x2 + 1

Another Example

Example
x3 + 2x2 + 4
x→∞ 3x2 + 1

Answer
The limit is ∞.

Solution
Solu on
x3 + 2x2 + 4 x3 (1 + 2/x + 4/x3 ) 1 + 2/x + 4/x3
= =x· , so
3x2 + 1 x2 (3 + 1/x2 ) 3 + 1/x2
( ) ( ) 1
x3 + 2x2 + 4 1 + 2/x + 4/x3
lim = lim x · = lim x · = ∞
x→∞ 3x2 + 1 x→∞ 3 + 1/x2 x→∞ 3

Yet Another Example
Example
Find
2x3 + 3x + 1
lim
x→∞ 4x3 + 5x2 + 7

if it exists.
A does not exist
B 1/2
C 0
D ∞

Solution
Solu on
2x3 + 3x + 1 x3 (2 + 3/x2 + 1/x3 )
= 3 , so
4x3 + 5x2 + 7 x (4 + 5/x + 7/x3 )
2x3 + 3x + 1 2 + 3/x2 + 1/x3 2+0+0 1
lim 3 2+7
= lim = =
x→∞ 4x + 5x x→∞ 4 + 5/x + 7/x3 4+0+0 2

Upshot of the last three examples
Upshot
When finding limits of algebraic expressions at infinity, look at the
highest degree terms.
If the higher degree is in the numerator, the limit is ±∞.
If the higher degree is in the denominator, the limit is 0.
If the degrees are the same, the limit is the ra o of the
top-degree coefficients.

Still Another Example
Example
Find √
3x4 + 7
lim
x→∞ x2 + 3

Still Another Example
√ √ √
3x4 + 7 ∼ 3x4 = 3x2
Example
Find √
3x4 + 7 .
lim
x→∞ x2 + 3

Answer
√
The limit is 3.

Solution
Solu on

√ √ √
3x4 + 7 x4 (3 + 7/x4 ) x2 (3 + 7/x4 )
lim = lim 2 = lim 2
x→∞ x2 + 3 x→∞ x (1 + 3/x2 ) x→∞ x (1 + 3/x2 )
√ √
(3 + 7/x4 ) 3+0 √
= lim = = 3.
x→∞ 1 + 3/x2 1+0

Rationalizing to get a limit
Example
(√ )
Compute lim 4x 2 + 17 − 2x .
x→∞

Rationalizing to get a limit
Example
(√ )
Compute lim 4x 2 + 17 − 2x .
x→∞

Solu on
This limit is of the form ∞ − ∞, which we cannot use. So we
ra onalize the numerator (the denominator is 1) to get an
expression that we can use the limit laws on.
(√ ) (√ ) √4x2 + 17 + 2x
lim 4x2 + 17 − 2x = lim + 17 − 2x · √
4x2
x→∞ x→∞ 4x2 + 17 + 2x
(4x + 17) − 4x
2 2
17
= lim √ = lim √ =0
x→∞ 4x2 + 17 + 2x x→∞ 4x2 + 17 + 2x

Kick it up a notch
Example
(√ )
Compute lim 4x 2 + 17x − 2x .
x→∞

Kick it up a notch
Example
(√ )
Compute lim 4x 2 + 17x − 2x .
x→∞

Solu on
Same trick, diﬀerent answer:
(√ )
lim 4x2 + 17x − 2x
x→∞
(√ ) √4x2 + 17x + 2x
= lim 4x2 + 17x − 2x · √
x→∞ 4x2 + 17x + 2x
(4x2 + 17x) − 4x2 17x 17 17
= lim √ = lim √ = lim √ =
x→∞ 4x2 + 17x + 2x x→∞ 4x2 + 17x + 2x x→∞ 4 + 17/x + 2 4

Summary

Inﬁnity is a more complicated concept than a single number.
There are rules of thumb, but there are also excep ons.
Take a two-pronged approach to limits involving inﬁnity:
Look at the expression to guess the limit.
Use limit rules and algebra to verify it.

Lesson 6: Limits Involving Infinity (slides)

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Lesson 6: Limits Involving Infinity (slides)