Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
Limits Involving Infinity
I. Infinite Limits
DEFINITION: The notation
lim f (x) = ∞
x→a
means that the values of f (x) can be made arbitrary large (as large as we like) by taking x
sufficiently close to a (on either side of a) but not equal to a.
Similarly,
lim f (x) = −∞
x→a
means that the values of f (x) can be made as large negative as we like by taking x sufficiently
close to a (on either side of a) but not equal to a.
Similar definitions can be given for the one-sided infinite limits
lim f (x) = ∞
x→a+
lim f (x) = ∞
lim f (x) = −∞
x→a+
x→a−
lim f (x) = −∞
x→a−
EXAMPLES:
1
1
1
, lim− 2 , lim+ 2 D.N.E., moreover
2
x→0 x
x→0 x
x→0 x
1
1
1
lim 2 = lim− 2 = lim+ 2 = ∞
x→0 x
x→0 x
x→0 x
1. The limits lim
x
±
±
±
±
±
0.1
0.01
0.001
0.0001
0.00001
f (x)
100
10000
1000000
100000000
10000000000
1
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
1
1
and lim+
D.N.E., moreover
x→5 5 − x
x→5 5 − x


WORK:
1


=
lim−
=∞
1
1
+“NOT SMALL”
x→5 5 − x
=
=
= +“BIG”
5 − 4.99
0.01
+“SMALL”
2. The limits lim−
and
lim+
x→5
Therefore lim

1

=
5−x
x→5

WORK:

 = −∞
1
1
+“NOT SMALL”
=
=
= −“BIG”
5 − 5.01
−0.01
−“SMALL”
1
D.N.E. and neither ∞ nor −∞ (see the Figure below (left)).
5−x
2x + 1
2x + 1
and lim +
D.N.E., moreover
x→−7
x→−7
x+7
x+7


WORK:
2x + 1 

lim −
=  2(−7.01) + 1
=∞
−13
−“NOT
SMALL”
x→−7
x+7
≈
=
= +“BIG”
−7.01 + 7
−0.01
−“SMALL”
3. The limits lim −
and
lim +
x→−7


WORK:
2x + 1 

=  2(−6.99) + 1
 = −∞
−13
−“NOT
SMALL”
x+7
≈
=
= −“BIG”
−6.99 + 7
0.01
+“SMALL”
2x + 1
D.N.E. and neither ∞ nor −∞ (see the Figure above (right)).
x→−7 x + 7
Therefore lim
2
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
REMARK: Recall that even if the denominator of a function f goes to 0, it does not necessary
mean that the limit of f is ∞ or −∞. In fact,
x2 − x − 2
x2 − x − 2
0
lim−
= lim+
=
x→2
x→2
x−2
0
x−2
x2 − x − 2
x→2
x−2
= lim
(x − 2)(x + 1)
x→2
x−2
= lim
= lim (x + 1) = 3
x→2
DEFINITION: The line x = a is called a vertical asymptote of the curve y = f (x) if at least
one of the following statements is true:
x→a
lim f (x) = ∞
x→a−
lim f (x) = −∞
x→a−
x→a
lim f (x) = ∞
x→a+
lim f (x) = −∞
x→a+
EXAMPLES:
1. Find all vertical asymptotes of f (x) = −
x2
8
.
−4
lim f (x) = ∞
lim f (x) = −∞
Solution: There are potential vertical asymptotes where x2 − 4 = 0, that is where x = ±2. In
fact, we have


WORK:
−8


=
lim − 2
 = −∞
−8
−8
−8
−“NOT
SMALL”
x→−2 x − 4
≈
=
=
=
−“BIG”
(−2.01)2 − 4
4.04 − 4
0.04
+“SMALL”
and


WORK:
−8


=
lim+ 2
 = −∞
−8
−8
−8
−“NOT SMALL”
x→2 x − 4
≈
=
=
=
−“BIG”
(2.01)2 − 4
4.04 − 4
0.04
+“SMALL”
This shows that the lines x = ±2 are the vertical asymptotes of f.
3
Section 1.6 Limits Involving Infinity
2. Find all vertical asymptotes of f (x) =
2010 Kiryl Tsishchanka
x2 − 1
.
x+1
Solution: There is a potential vertical asymptote where x + 1 = 0, that is where x = −1.
Moreover, if x = −1 is a vertical asymptote, it is the only one. However,
(x − 1)(x + 1)
x2 − 1
= lim
= lim (x − 1) = −2
lim
x→−1
x→−1
x→−1 x + 1
x+1
x2 − 1
= −2. This shows that f does not have vertical asymptotes (see
x→−1 x + 1
the Figure below (left)).
and therefore lim ±
3. Find all vertical asymptotes of f (x) =
x2 − 3x + 2
.
x2 − 4x + 3
Solution: There are potential vertical asymptotes where x2 − 4x + 3 = 0, that is where x = 1, 3.
However,
x2 − 3x + 2
(x − 1)(x − 2)
x−2
1
lim 2
= lim
= lim
=
x→1 x − 4x + 3
x→1 (x − 1)(x − 3)
x→1 x − 3
2
and therefore lim±
hand,
lim+
x→3
x→1
x2 − 3x + 2
1
= . Hence x = 1 is not a vertical asymptote. On the other
2
x − 4x + 3
2
x2 − 3x + 2
(x − 1)(x − 2)
= lim+
2
x − 4x + 3 x→3 (x − 1)(x − 3)


WORK:
x−2 

=  3.01 − 2
= lim+
=∞
1.01
+“NOT
SMALL”
x→3 x − 3
=
=
= +“BIG”
3.01 − 3
0.01
+“SMALL”
This shows that the line x = 3 is the only vertical asymptote of f (see the Figure above (right)).
4
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
4. Find all vertical asymptotes of f (x) = cot 2x.
Solution: Because
cos 2x
sin 2x
there are potential vertical asymptotes where sin 2x = 0. In fact, since sin 2x → 0+ as x → 0+
and sin 2x → 0− as x → 0− , whereas cos 2x is positive (and not near 0) when x is near 0, we
have
lim+ cot 2x = ∞ and lim− cot 2x = −∞
cot 2x =
x→0
x→0
This shows that the line x = 0 is a vertical asymptote. Similar
reasoning shows that the lines x = nπ/2, where n is an integer, are all vertical asymptotes of f (x) = cot 2x. The graph
confirms that.
II. 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.
Similarly, the notation
lim f (x) = L
x→−∞
means that the values of f (x) can be made arbitrary close to L by taking x sufficiently large
negative.
DEFINITION: The line y = L is called a horizontal asymptote of the curve y = f (x) if
either
lim f (x) = L or
lim f (x) = L
x→∞
x→−∞
5
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
EXAMPLE: Find the horizontal asymptote of f (x) =
x2 − 1
.
x2 + 1
Solution: We have
x2 − 1 h ∞ i A
=
lim
= lim
x→±∞ x2 + 1
x→±∞
∞
x2
x2 −1
2
A
x2
= lim xx2
x2 +1
x→±∞
x2
x2
−
+
1
x2 A
= lim
1
x→±∞
x2
1−
1+
1
x2 C
1 =
x2
1−0
=1
1+0
x2 − 1
It follows that y = 1 is the horizontal asymptote of f (x) = 2
. The graph below confirms
x +1
that:
EXAMPLE: Find all horizontal asymptotes of f (x) = √
x
.
x2 + 1
Solution: We have
h∞i
x
A
lim √
=
= lim
x→∞
x→∞
∞
x2 + 1
√x
A
x2
√
= lim
2
x +1
x→∞
√
x2
x
qx
x2 +1
x2
x→∞
A
= lim q
x→∞
and
h∞i
x
A
=
lim √
= lim
2
x→−∞
x→−∞
∞
x +1
√x
A
x2
√
= lim
2
+1
x
x→−∞
√
x2
x
q−x
x2 +1
x2
1
A
= lim q
x2
x2
+
1
x→−∞
= lim q
6
1
x2
x2
x2
+
−1
1+
It follows that y = ±1 are the horizontal asymptotes of f (x) = √
confirms that:
=√
1
=1
1+0
−1
A
x→−∞
C
1+
= lim q
A
1
x2
1
x2
C
1
x2
=√
−1
= −1
1+0
x
. The graph below
+1
x2
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
EXAMPLES:
2x3 − x + 5 h ∞ i A
=
= lim
1. lim 3
x→∞
x→∞ x + x2 − 1
∞
C
=
2x3
2x3 −x+5
− xx3 + x53 A
A
x3
x3
=
lim
= lim
2
3
x3 +x2 −1
x→∞ x + x − 1
x→∞
3
3
3
3
x
x
x
x
2−0+0
=2
1+0−0
3x + 3x2 − 7 h ∞ i A
2. lim
=
= lim
x→±∞ x + 1 − 5x2
x→±∞
∞
C
=
3x
3x+3x2 −7
2
A
x2
= lim xx
x+1−5x2
x→±∞
x2
x2
+
+
3x2
3
− x72 A
x2
x
lim
2 =
1
x→±∞ 1
− 5x
x
x2
x2
3
0+3−0
=−
0+0−5
5
√
√
h∞i
4x1/3 − x1/2 − 1 A
43x− 2x−1
A
√
= lim
=
lim
=
3. lim √
x→∞
x→∞ 2x1/3 − 9x1/2 + 1
x→∞ 2 3 x − 9 2 x + 1
∞
A
=
A
2 − x12 + x53
1 + x1 − x13
1/2
1
4x1/3
− xx1/2 − x1/2
A
x1/2
lim 1/3 9x1/2
= lim
1
x→∞ 2x
x→∞
− x1/2 + x1/2
x1/2
4
x1/6
2
x→∞ 1/6
x
= lim
−1−
−9+
1
x1/2 C
1 =
x1/2
4x1/3−1/2 − 1 −
2x1/3−1/2 − 9 +
A
x→−∞
4x−1/6 − 1 −
2x−1/6 − 9 +
0−1−0
−1
1
=
=
0−9+0
−9
9
7x2 + 10x + 20 h ∞ i A
4. lim
=
= lim
x→−∞ x3 − 10x2 − 1
x→−∞
∞
= lim
1
x1/2 A
lim
1 = x→∞
x1/2
4x1/3 −x1/2 −1
x1/2
2x1/3 −9x1/2 +1
x1/2
7x2 +10x+20
7x2
+ 10x
A
x3
x3
x3
= lim x3 10x
2
x3 −10x2 −1
x→−∞
− x3
x3
x3
+
−
20
x3
1
x3
+ x102 + x203 C 0 + 0 + 0
=
=0
1−0−0
− x13
1 − 10
x
7
x
or
7x2 + 10x + 20 h ∞ i A
lim
=
= lim
x→−∞ x3 − 10x2 − 1
x→−∞
∞
7x2
7x2 +10x+20
+ 10x
A
x2
x2
x2
=
lim
3
2
x3 −10x2 −1
x
10x
x→−∞
− x2
x2
x2
+ x202 C
7 + 10
7+0+0 C
7
C
x
= lim
= lim
=0
= lim
x→−∞ x − 10 − 0
x→−∞ x − 10
x→−∞ x − 10 − 12
x
A
5.
11x5 + 1
x→−∞ 4 − x4
lim
7
+
−
20
x2
1
x2
1
x1/2
1
x1/2
+ 3 − x72
+ x12 − 5
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
11x5 +1
11x5
+ x14
11x5 + 1 h ∞ i A
A
x4
x4
5. lim
=
= lim 4−x4 = lim 4
4
x→−∞ 4 − x4
x→−∞
x→−∞
∞
− xx4
x4
x4
11x + x14 C −∞
A
= ∞ (D.N.E)
=
= lim
4
x→−∞
−1
−
1
4
x
or
11x5
11x5 +1
+ x15
11x5 + 1 h ∞ i A
A
x5
x5
lim
=
lim
=
=
lim
4
4
x→−∞ 4 − x4
x→−∞ 4 − x
x→−∞ 4−x
∞
x5
x5
x5
11 + x15 C 11
A
= lim 4
= + = ∞ (D.N.E)
x→−∞ 5 − 1
0
x
x
h∞i
x2 + 2x + 5
A
=
6. lim √
= lim
4
x→∞
x→∞
∞
2 + x + 7x
x2
x2
A
= lim q
x→∞
2
x4
+
2x
x2
+
+
x
x4
+
5
x2
7x4
x4
x2 +2x+5
√
A
x4
√
= lim
4
2+x+7x
x→∞
√
x4
1+
= lim q
A
x→∞
2
x4
2
x
+
+
1
x3
5
x2
x2 +2x+5
x2
q
2+x+7x4
x4
1
C 1+0+0
=√
=√
0+0+7
7
+7
v
s
s
u
√
2x5 − 1 A u
2x5 − 1 h ∞ i A
2x5 − 1 C
=
= t lim
7. lim √
= lim
lim
=
x→∞
x→∞
x→∞ 3x5 + 2
x→∞
∞
3x5 + 2
3x5 + 2
s
s
r
r
1
2x5
−
2 − x15 C
2−0
2
A
x5
x5 A
=
=
=
lim 3x5
lim
=
2
x→∞
x→∞ 3 + 5
3+0
3
+ x25
x
x5
2x5 −1
x5
3x5 +2
x5
REMARK: For more examples of this type, see Appendix I.
h i
∞ A
x+3
C
8. lim sin
=
= sin lim
= sin lim
x→∞
x→∞ 6x2 − 5
x→∞
∞
1
+ x32 C
0
0+0
A
x
= sin
= sin 0 = 0
= sin
= sin lim
5
x→∞ 6 − 2
6−0
6
x
9.
lim (
x→∞
√
x2
x+3
6x2 − 5
A
+ 4 − x) = [∞ − ∞] = lim
x→∞
√
x+3
x2
6x2 −5
x2
!
√
√
x2 + 4 − x A
( x2 + 4 − x)( x2 + 4 + x)
√
= lim
x→∞
1
1 · ( x2 + 4 + x)
√
x2 + 4 − x2 A
4
( x2 + 4)2 − x2 A
C
= lim √
= lim √
=0
= lim √
2
2
2
x→∞
x→∞
x +4+x
x + 4 + x x→∞ x + 4 + x
A
10.
A
= sin
x
+ x32
x2
lim 6x
2
x→∞
− x52
x2
√
lim ( x2 + 4 − x)
x→−∞
8
!
Section 1.6 Limits Involving Infinity
10.
2010 Kiryl Tsishchanka
√
lim ( x2 + 4 − x) = [∞ + ∞] = ∞ (D.N.E)
x→−∞
REMARK: For more examples of this type, see Appendix II.
11.
12.
13.
14.
15.
lim (x −
x→∞
√
x)
lim (x3 − x8 )
x→∞
lim (x3 + x8 )
x→∞
lim sin x
x→∞
lim
x→∞
1 + sin 5x
√
1+x
9
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
√
√
√
√
(x − x)(x + x) A
x2 − ( x)2
x− x A
√
√
11. lim (x − x) = [∞ − ∞] = lim
= lim
= lim
x→∞
x→∞
x→∞
x→∞
1
1 · (x + x)
x+ x
x2
x2 −x
− xx A
x − 1 C h∞i
x2 − x A
A
A
x
x
√ = lim
√ = lim x+√x = lim
=
= ∞ (D.N.E)
= lim
x→∞ x + x
x→∞ 1 + √1
x→∞ x +
1
x x→∞
x
√
A
x
x
x
or
lim (x −
x→∞
12.
13.
14.
15.
√
√ √
√ A
√ √
C
A
x) = [∞ − ∞] = lim ( x · x − 1 · x) = lim [ x( x − 1)] = [∞ · ∞] = ∞
x→∞
x→∞
(D.N.E)
A
C
lim (x3 − x8 ) = [∞ − ∞] = lim x3 (1 − x5 ) = [∞ · (−∞)] = −∞ (D.N.E)
x→∞
x→∞
lim (x3 + x8 ) = [∞ + ∞] = ∞ (D.N.E)
x→∞
lim sin x D.N.E
x→∞
lim
x→∞
1 + sin 5x
√
=0
1+x
Solution: We first note that
0 ≤ 1 + sin 5x ≤ 2
√
Dividing all three parts of this inequality by 1 + x, we get
1 + sin 5x
2
0≤ √
≤√
1+x
1+x
Since
lim √
x→∞
it follows that
2
=0
1+x
1 + sin 5x
√
=0
x→∞
1+x
lim
by the Squeeze Theorem.
10
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
Appendix I
√
3
x+5
.
1. Find lim √
x→∞
x+7
Solution: We have
√
3
x + 5 h∞i A
lim √
=
= lim
x→∞
x→∞
∞
x+7
√
3
x+5
√
x
√
x+7
√
x
q
3
A
= lim q
x→∞
=
√
1
2
x=x =x
x+5
x3/2 A
x+7
x
q
x
3
= lim
x→∞
x3/2
q
x
x
3 1
·
2 3
√
3
x+5
√
3 3/2
x
√
x+7
x→∞ √
x
3 31 √
3
A
3/2
= x
= x2
= lim
5
+
3
x3/2 A
= lim
x→∞
7
x
+
q
1
x1/2
+
q
1+
√
3
0+0
0
√
=
= =0
1
1+0
5
x3/2 C
7
x
√
6
3x2 + 4
.
2. Find lim √
x→∞ 9 1 − 2x3
Solution: We have
√
6
3x2 + 4 h ∞ i A
lim √
=
= lim
x→∞ 9 1 − 2x3
x→∞
∞
√
6
2 +4
3x
√
9 3
x
√
9
3
1−2x
√
9 3
x
q
6
A
= lim q
x→∞
9
=
n√
9
x3 = x = x = x
3x2 +4
x2
A
1−2x3
x3
1
3
3
9
q
6
= lim q
x→∞
9
11
3x2
x2
1
x3
+
−
= x
1
2 6
√
6
2 +4
3x
√
6 2
x
√
9
3
1−2x
x→∞
√
9 3
x
o
√
6
A
= x2 = lim
q
√
6
6
6
3 + x42 C √
3+0
3
√
=
−
= √
= lim q
9
9
x→∞ 9 1
0−2
2
−2
x3
4
x2 A
2x3
x3
2· 16
Section 1.6 Limits Involving Infinity
2010 Kiryl Tsishchanka
Appendix II
√
1. Find lim [x( x2 + 4 − x)].
x→∞
Solution: Since
√
lim ( x2 + 4 − x) = 0
x→∞
√
by Example 9 from page 8, it follows that lim [x( x2 + 4 − x)] is ∞ · 0 type of an indeterminate
x→∞
form. We have
√
√
√
√
x( x2 + 4 − x)( x2 + 4 + x)
x( x2 + 4 − x) A
A
2
√
= lim
lim [x( x + 4 − x)] = [∞ · 0] = lim
x→∞
x→∞
x→∞
1
1 · ( x2 + 4 + x)
√
x[x2 + 4 − x2 ]
x[( x2 + 4)2 − x2 ] A
A
√
= lim √
= lim
x→∞
x→∞
x2 + 4 + x
x2 + 4 + x
4x
h∞i
4
4x
A
A
A
= lim √ 2
= lim √
=
= lim √ 2 x
x→∞ x +4 + x
x→∞
x→∞ x +4+x
∞
x2 + 4 + x
x
x
x
A
4
= lim
x→∞
+1
4
A
= lim q
4
A
√
x2 +4
x→∞ √
x2
1+
= lim q
x→∞
C
4
x2
+1
=√
x2 +4
x2
4
A
+1
= lim q
x→∞
x2
x2
+
4
x2
+1
4
=2
1+0+1
√
2. Find lim [x( x2 + 4 + x)].
x→−∞
Solution: We have
lim [x(
x→−∞
√
x2
√
√
√
x( x2 + 4 + x)( x2 + 4 − x)
x( x2 + 4 + x) A
√
= lim
+ 4 + x)] = [∞ · 0] = lim
x→−∞
x→−∞
1
1 · ( x2 + 4 − x)
√
x[x2 + 4 − x2 ]
x[( x2 + 4)2 − x2 ] A
A
√
= lim √
= lim
x→−∞
x→−∞
x2 + 4 − x
x2 + 4 − x
4x
h∞i
4
4x
A
A
A
= lim √ 2
= lim √
=
= lim √ 2 x
x→−∞ x +4 −
x→−∞
x→−∞ x +4−x
∞
x2 + 4 − x
x
x
A
A
= lim
x→−∞
√
4
x2 +4
√
− x2
x→−∞
−1
x→−∞
4
A
= lim
−
q
4
A
= lim
1+
12
C
=
4
x2
−1
−
q
x2 +4
x2
x→−∞
4
= −2
− 1+0−1
√
4
A
= lim
−1
x
x
−
q
x2
x2
+
4
x2
−1