Lesson 5: Limits Involving Infinity

Section 1.6
Limits involving Inﬁnity

V63.0121.002.2010Su, Calculus I

New York University

May 20, 2010

Announcements

Oﬃce Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here)
Quiz 1 Thursday on 1.1–1.4

Announcements

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

“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.

“Large” takes the place of x
“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
x→0 x



y

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


Finding limits at trouble spots

Example
Let
x2 + 2
f (x) =
x 2 − 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) =
x 2 − 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

− 0 +
(x − 1)
1

So

Use the number line

− 0 +
(x − 1)
1
− 0 +
(x − 2)
2

So

Use the number line

− 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x 2 + 2)

So

Use the number line

− 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x 2 + 2)

f (x)
1 2

So

Use the number line

− 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x 2 + 2)
+
f (x)
1 2

So

Use the number line

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

So
lim f (x) = +∞
x→1−

Use the number line

− 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x 2 + 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
+
(x 2 + 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
+
(x 2 + 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
+
(x 2 + 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+

Use the number line

− 0 +
(x − 1)
1
− 0 +
(x − 2)
2
+
(x 2 + 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

y

x
−1 1 2 3


Limit Laws (?) with inﬁnite limits

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

∞+∞=∞

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

−∞ − ∞ = −∞


Rules of Thumb with inﬁnite limits

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

∞+∞=∞

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

−∞ − ∞ = −∞



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

L+∞=∞
L − ∞ = −∞


Kids, don’t try this at home!

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.

−∞ if L > 0
L · (−∞) =
∞ if L < 0.


Multiplying infinite limits

The product of two infinite limits is infinite.

∞·∞=∞
∞ · (−∞) = −∞
(−∞) · (−∞) = ∞


Dividing by Infinity

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.



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 sin(1/x)
x→0 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


Limits at infinity

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.


Limits at infinity

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→−∞


Limits at infinity

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→∞ x n
1
lim =0
x→−∞ x n


Using the limit laws to compute limits at ∞

Example
Find
2x 3 + 3x + 1
lim
x→∞ 4x 3 + 5x 2 + 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
2x 3 + 3x + 1 x 3 (2 + 3/x 2 + 1/x 3 )
= 3
4x 3 + 5x 2 + 7 x (4 + 5/x + 7/x 3 )
2x 3 + 3x + 1 2 + 3/x 2 + 1/x 3
lim 3 + 5x 2 + 7
= lim
x→∞ 4x x→∞ 4 + 5/x + 7/x 3
2+0+0 1
= =
4+0+0 2


Solution
Factor out the largest power of x from the numerator and denominator.
We have
2x 3 + 3x + 1 x 3 (2 + 3/x 2 + 1/x 3 )
= 3
4x 3 + 5x 2 + 7 x (4 + 5/x + 7/x 3 )
2x 3 + 3x + 1 2 + 3/x 2 + 1/x 3
lim 3 + 5x 2 + 7
= lim
x→∞ 4x x→∞ 4 + 5/x + 7/x 3
2+0+0 1
= =
4+0+0 2

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


Another Example

Example
x
Find lim
x→∞ x2 + 1


Another Example

Example
x
Find lim
x→∞ x2 + 1

Answer
The limit is 0.


Solution
Again, 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/x 2) x 1 + 1/x 2
x 1 1 1 1
lim 2 = lim = lim · lim
x→∞ x + 1 x→∞ x 1 + 1/x 2 x→∞ x x→∞ 1 + 1/x 2
1
=0· = 0.
1+0


Another Example

Example
x
Find lim
x→∞ x2 + 1

Answer
The limit is 0.
y

x


Another Example

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 heuristic deﬁnitions of asymptote.


Solution
Again, 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/x 2) x 1 + 1/x 2
x 1 1 1 1
lim 2 = lim = lim · lim
x→∞ x + 1 x→∞ x 1 + 1/x 2 x→∞ x x→∞ 1 + 1/x 2
1
=0· = 0.
1+0

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


Another Example

Example
Find

√
3x 4 + 7
lim
x→∞ x2 + 3


Another Example

Example
Find

√ √
3x 4 + 7 ∼ 3x 4 = 3x 2
√
3x 4 + 7
lim
x→∞ x2 + 3

Answer
√
The limit is 3.


Solution

√
3x 4 + 7 x 4 (3 + 7/x 4 )
lim = lim
x→∞ x2 + 3 x→∞ x 2 (1 + 3/x 2 )

x 2 (3 + 7/x 4 )
= lim
x→∞ x 2 (1 + 3/x 2 )

(3 + 7/x 4 )
= lim
x→∞ 1 + 3/x 2
√
3+0 √
= = 3.
1+0


Example
Compute lim 4x 2 + 17 − 2x .
x→∞


Example
Compute lim 4x 2 + 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.
√
4x 2 + 17 + 2x
lim 4x 2 + 17 − 2x = lim 4x 2 + 17 − 2x · √
x→∞ x→∞ 4x 2 + 17 + 2x
(4x 2 + 17) − 4x 2
= lim √
x→∞ 4x 2 + 17 + 2x
17
= lim √ =0
x→∞ 4x 2 + 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→∞

Solution
Same trick, diﬀerent answer:

lim 4x 2 + 17x − 2x
x→∞
√
4x 2 + 17x + 2x
= lim 4x 2
+ 17x − 2x · √
x→∞ 4x 2 + 17x + 2x
(4x 2 + 17x) − 4x 2
= lim √
x→∞ 4x 2 + 17x + 2x
17x 17 17
= lim √ = lim =
x→∞ 4x 2 + 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 5: Limits Involving Infinity

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