Section 11.2 One-sided Limits and Limits Involving Infinity
One-sided Limits
DEFINITION: We write
lim f (x) = L
x→a−
and say the left-hand limit of f (x) as x approaches a is equal to L if we can make the values of
f (x) arbitrary close to L by taking x to be sufficiently close to a and x less than a.
Similarly, we write
lim f (x) = L
x→a+
and say the right-hand limit of f (x) as x approaches a is equal to L if we can make the values of
f (x) arbitrary close to L by taking x to be sufficiently close to a and x greater than a.
IMPORTANT:
lim f (x) = L if and only if
lim f (x) = L and
x→a
x→a−
lim f (x) = L
x→a+
EXAMPLE: The graph of a function g is shown below.
Then
(a) lim− g(x) = 3
(b) lim+ g(x) = 1
(c) lim g(x) D.N.E.
(d) lim− g(x) = 2
(e) lim+ g(x) = 2
(f) lim g(x) = 2
x→2
x→5
x→5
x→5
EXAMPLE: Let
f (x) =
Then
x2
if x ≤ 2
x + 1 if x > 2
DSP
lim− f (x) = lim− x2 = 22 = 4 and
x→2
x→2
x→2
x→2
DSP
lim+ f (x) = lim+ (x + 1) = 2 + 1 = 3.
x→2
x→2
Since lim− f (x) 6= lim+ f (x), it follows that lim f (x) does not exist.
x→2
EXAMPLE: Let
Then
x→2
x→2

x+2


 x − 1 if x ≤ −1
g(x) =
3
if − 1 < x ≤ 0
 x

 √
x
if x > 0
1
1
−1 + 2
=−
(b) lim + g(x) = (−1)3 = −1 (c) lim g(x) D.N.E. (d) g(−1) = −
x→−1
x→−1
x→−1
−1 − 1
2
2
√
3
3
(g) lim g(x) = 0
(h) g(0) = 0 = 0
(e) lim− g(x) = 0 = 0
(f) lim+ g(x) = 0 = 0
(a)
lim − g(x) =
x→0
x→0
x→0
1
EXAMPLE: Find each of the given limits.
√
√
(a) lim+ 4 − x2 and lim− 4 − x2
x→2
x→2
√
Solution: Since f (x) = 4 − x2 is not defined when x > 2, the right-hand limit (which requires that
x > 2) does not exist. For the left-hand limit, write the square root in exponential form and apply the
appropriate limit properties
√
lim− 4 − x2 = lim− (4 − x2 )1/2 = [ lim− (4 − x2 )]1/2 = (4 − 22 )1/2 = 01/2 = 0
x→2
x→2
x→2
√
2
(b) lim+ [ x − 3 + x + 1]
x→3
Solution: We have
√
lim+ [ x − 3 + x2 + 1] = lim+ [(x − 3)1/2 + x2 + 1]
x→3
x→3
= lim+ (x − 3)1/2 + lim+ x2 + lim+ 1
x→3
x→3
x→3
= [ lim+ (x − 3)]1/2 + lim+ x2 + lim+ 1 = (3 − 3)1/2 + 32 + 1 = 0 + 9 + 1 = 10
x→3
x→3
x→3
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−
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
EXAMPLES:
lim f (x) = ∞
x→a+
lim f (x) = ∞
lim f (x) = −∞
x→a+
lim f (x) = −∞
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
0.000001
f (x)
100
10000
1000000
100000000
10000000000
1000000000000
It follows that x = 0 is the vertical asymptote of the function f (x) =
1
.
x2
1
1
and lim+
D.N.E., moreover
x→5 5 − x
x→5 5 − x


WORK:
1


lim−
=
=∞
1
+“NOT
SMALL”
1
x→5 5 − x
=
=
= +“BIG”
5 − 4.99
0.01
+“SMALL”
2. The limits lim−
and
lim+
x→5
It follows that

1

=
5−x

WORK:

 = −∞
1
1
+“NOT SMALL”
=
=
= −“BIG”
5 − 5.01
−0.01
−“SMALL”
(a) x = 5 is the vertical asymptote of the function f (x) =
1
D.N.E. and neither ∞ nor −∞.
x→5 5 − x
(b) lim
3
1
;
5−x
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
x→∞
EXAMPLE: Find the horizontal asymptote of f (x) =
lim f (x) = L
x→−∞
x2 − 1
.
x2 + 1
Solution: We have
x2 − 1
2
x −1
A
A
lim 2
=
= lim 2x
= lim
x→±∞ x + 1
x→±∞ x + 1
x→±∞
∞
2
x
2
h∞i
1
x2
− 2
1−
2
A
x
x =
lim
x→±∞
x2
1
1+
+ 2
2
x
x
It follows that y = 1 is the horizontal asymptote of f (x) =
4
1
C 1−0
x2 =
=1
1
1+0
x2
x2 − 1
. The graph below confirms that:
x2 + 1
EXAMPLES:
x
5
2x3
− 3+ 3
3
x
x
x
x3 x2
1
+
−
x3 x3 x3
1
5
2− 2 + 3
C 2−0+0
A
x
x =
=2
= lim
1
1
x→±∞
1+0−0
1+ − 3
x x
2x3 − x + 5
It follows that y = 2 is the horizontal asymptote of the function f (x) = 3
.
x + x2 − 1
2x3 − x + 5
3
2x − x + 5
A
A
1. lim 3
=
= lim
= lim 3 x 2
2
x→±∞ x + x − 1
x→±∞
x→±∞ x + x − 1
∞
3
x
3
h∞i
7
3x 3x2
+ 2 − 2
2
x
x
x
x
1
5x2
+
− 2
x2 x2
x
3
7
+3− 2
3
C 0+3−0
A
x =
= lim x
=−
1
x→±∞ 1
0+0−5
5
+ 2 −5
x x
3x + 3x2 − 7
3
.
It follows that y = − is the horizontal asymptote of the function f (x) =
5
x + 1 − 5x2
3x + 3x2 − 7
3x + 3x − 7
A
A
x2
=
= lim
= lim
2. lim
2
2
x→±∞ x + 1 − 5x
x→±∞
x→±∞ x + 1 − 5x
∞
2
x
2
h∞i
7x2 + 10x + 20
7x + 10x + 20
A
A
x3
=
= lim
= lim
3. lim
3
3
2
x→±∞ x − 10x2 − 1
x→±∞
x→±∞ x − 10x − 1
∞
3
x
2
7x2 10x
+ 3 +
x3
x
x3 10x2
− 3 −
x3
x
7 10 20
+ 2+ 3
A
x
x
x
= lim
1
10
x→±∞
− 3
1−
x
x
h∞i
It follows that y = 0 is the horizontal asymptote of the function f (x) =
20
x3
1
x3
C
=
0+0+0
=0
1−0−0
7x2 + 10x + 20
.
x3 − 10x2 − 1
11x5 + 1
11x5
1
+ 4
4
4
11x5 + 1
A
A
x
x
=
= lim
= lim x
4. lim
x→−∞ 4 − x4
x→−∞
x→−∞ 4 − x4
x4
4
∞
−
x4
x4 x4
h∞i
1
−∞
C
x4 =
= ∞ (D.N.E)
4
−1
−
1
x4
11x +
A
= lim
x→−∞
11x5 + 1
11x5 + 1
=
−∞.
It
follows
that
the
function
f
(x)
=
x→∞ 4 − x4
4 − x4
does not have horizontal asymptotes.
In the same way one can show that lim
5