Lesson 6: Limits Involving ∞ (Section 41 slides)

Section 1.6
Limits involving Infinity

V63.0121.041, Calculus I

New York University

September 22, 2010

Announcements

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

. . . . . .

Announcements

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

. . . . . .

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

Objectives

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

. . . . . .


Recall the definition of limit

Definition
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.

. . . . . .


Recall the unboundedness problem
1
Recall why lim+ doesn’t exist.
x→0 x

y
.

.?.
L

. x
.

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


Outline

Infinite Limits
Vertical Asymptotes
Infinite Limits we Know
Limit “Laws” with Infinite Limits
Indeterminate Limit forms

Limits at ∞
Algebraic rates of growth
Rationalizing to get a limit

. . . . . .


Infinite Limits

Definition
The notation
y
.
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 to a but not
equal to a.
. x
.

. . . . . .


Infinite Limits

Definition
The notation
y
.
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 to a but not
equal to a.
. x
.
“Large” takes the place of
“close to L”.

. . . . . .


Negative Infinity

Definition
The notation
lim f(x) = −∞
x→a

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

. . . . . .


Negative Infinity

Definition
The notation
lim f(x) = −∞
x→a

means that the values of f(x) can be made arbitrarily large negative (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 (negative) number.

. . . . . .


Vertical Asymptotes

Definition
The line x = a is called a vertical 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 =∞
x→0 x2 .

.

.
. . . . . .


Finding limits at trouble spots

Example
Let
x2 + 2
f(x) =
x2 − 3x + 2
Find lim f(x) and lim+ f(x) for each a at which f is not continuous.
x→a− x→a

. . . . . .


Finding limits at trouble spots

Example
Let
x2 + 2
f(x) =
x2 − 3x + 2
Find lim f(x) and lim+ f(x) for each a at which f is not continuous.
x→a− x→a

Solution
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

−
.. 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

−
.. 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

−
.. 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

−
.. 0
.. .
+
. x − 1)
(
1
.
−
. 0
.. .
+
. x − 2)
(
2
.
.
+
. x2 + 2)
(
.
+ . ∞ .. ∞ . . ∞ .. ∞
+ − − − + .
+
f
.(x)
1
. 2
.

So

. . . . . .


Use the number line

−
.. 0
.. .
+
. x − 1)
(
1
.
−
. 0
.. .
+
. x − 2)
(
2
.
.
+
. x2 + 2)
(
.
+ . ∞ .. ∞ . . ∞ .. ∞
+ − − − + .
+
f
.(x)
1
. 2
.

So
lim f(x) = +∞
x→1−

. . . . . .


Use the number line

−
.. 0
.. .
+
. x − 1)
(
1
.
−
. 0
.. .
+
. x − 2)
(
2
.
.
+
. x2 + 2)
(
.
+ . ∞ .. ∞ . . ∞ .. ∞
+ − − − + .
+
f
.(x)
1
. 2
.

So
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
.

So
lim f(x) = +∞ lim f(x) = −∞
x→1− x→2−
lim f(x) = −∞
x→1+

. . . . . .


Use the number line

−
.. 0
.. .
+
. x − 1)
(
1
.
−
. 0
.. .
+
. x − 2)
(
2
.
.
+
. x2 + 2)
(
.
+ . ∞ .. ∞ . . ∞ .. ∞
+ − − − + .
+
f
.(x)
1
. 2
.

So
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:
“As x → 1− , the numerator approaches 3, and the denominator
approaches 0 while remaining positive. 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

If lim f(x) = ∞ and lim g(x) = ∞,
x→a x→a

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

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

. . . . . .



L+∞=∞
.
L − ∞ = −∞

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

. . . . . .


Kids, don't try this at home!

Fact

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

. . . . . .



{
∞ if L > 0
. ·∞=
L
−∞ if L < 0.
Fact

limit is not 0. .

. . . . . .



{
∞ if L > 0
. ·∞=
L
−∞ if L < 0.
Fact

limit is not 0. . .

{
−∞ if L > 0
. · (−∞) =
L
∞ if L < 0.

. . . . . .


Multiplying infinite limits

Fact

The product of two infinite limits is infinite.

. . . . . .


Multiplying infinite limits

∞·∞=∞
. ∞ · (−∞) = −∞
(−∞) · (−∞) = ∞
Fact

The product of two infinite limits is infinite. .

. . . . . .


Dividing by Infinity

Fact

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

. . . . . .


Dividing by Infinity

L
. =0
∞

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

. . . . . .


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.

. . . . . .



L
Limits of the form are indeterminate. There is no rule for evaluating
0
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, even
x→0 x sin(1/x) 0
in the left- or right-hand sense. There are infinitely many vertical
asymptotes arbitrarily close to 0!

. . . . . .



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.
. . . . . .


Outline

Infinite Limits
Vertical Asymptotes
Limit “Laws” with Infinite Limits

Limits at ∞
Algebraic rates of growth

. . . . . .


Definition
Let f be a function 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.

. . . . . .


Definition

lim f(x) = L
x→∞


Definition
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→−∞

. . . . . .


Definition

lim f(x) = L
x→∞


Definition
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 positive integer. Then
1
lim =0
x→∞ xn
1
lim =0
x→−∞ xn

. . . . . .


Limit laws at infinity

Fact
Any limit law that concerns finite limits at a finite point a is still true if
the finite point is replaced by infinity.
That is, if lim f(x) = L and lim g(x) = M, then
x→∞ x→∞
lim (f(x) + g(x)) = L + M
x→∞
lim (f(x) − g(x)) = L − M
x→∞
lim cf(x) = c · L (for any constant c)
x→∞
lim f(x) · g(x) = L · M
x→∞
f(x) L
lim = (if M ̸= 0)
x→∞ g(x) M

. . . . . .


Using the limit laws to compute limits at ∞

Example
x
Find lim
x→∞ x2 +1

. . . . . .



Example
x
Find lim
x→∞ x2 +1

Answer
The limit is 0.
y
.

. x
.

. . . . . .


Solution

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

. . . . . .



Example
x
Find lim
x→∞ x2 +1

Answer
The limit is 0.
y
.

. x
.

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


Solution

Solution
We have
x x(1) 1 1
= 2 = ·
x2
+1 x (1 + 1/x2 ) x 1 + 1/x2
x 1 1 1 1
lim = lim = lim · lim
x→∞ x2 + 1 x→∞ x 1 + 1/x2 x→∞ x x→∞ 1 + 1/x2
1
=0· = 0.
1+0

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


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

Solution
We have
2x3 + 3x + 1 x3 (2 + 3/x2 + 1/x3 )
= 3
4x3 + 5x2 + 7 x (4 + 5/x + 7/x3 )
2x3 + 3x + 1 2 + 3/x2 + 1/x3
lim = lim
x→∞ 4x3 + 5x2 + 7 x→∞ 4 + 5/x + 7/x3
2+0+0 1
= =
4+0+0 2

. . . . . .


Solution

Solution
We have
2x3 + 3x + 1 x3 (2 + 3/x2 + 1/x3 )
= 3
4x3 + 5x2 + 7 x (4 + 5/x + 7/x3 )
2x3 + 3x + 1 2 + 3/x2 + 1/x3
lim = lim
x→∞ 4x3 + 5x2 + 7 x→∞ 4 + 5/x + 7/x3
2+0+0 1
= =
4+0+0 2

Upshot
When finding limits of algebraic expressions at infinity, look at the
highest degree terms.
. . . . . .


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

Solution

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

. . . . . .


Example
(√ )
Compute lim 4x2 + 17 − 2x .
x→∞

. . . . . .


Example
(√ )
Compute lim 4x2 + 17 − 2x .
x→∞

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


Kick it up a notch
Example
(√ )
Compute lim 4x2 + 17x − 2x .
x→∞

. . . . . .


Kick it up a notch
Example
(√ )
Compute lim 4x2 + 17x − 2x .
x→∞

Solution
Same trick, different answer:
(√ )
lim 4x2 + 17x − 2x
x→∞
(√ √ )
4x2 + 17 + 2x
= lim + 17x − 2x · √
4x2
x→∞ 4x2 + 17x + 2x
(4x2 + 17x) − 4x2
= lim √
x→∞ 4x2 + 17x + 2x
17x 17 17
= lim √ = lim √ =
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 exceptions.
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 ∞ (Section 41 slides)

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