Download Limits at Infinity

Differential Calculus 201-NYA-05
Vincent Carrier
Limits at Infinity
A limit at infinity is a limit of the form
lim f (x)
or
x→−∞
lim f (x).
x→∞
The result of a limit at infinity describes the way f behaves when x assumes arbitrarily
large negative and positive values.
It is fundamental to understand that
lim f (x) = 0
whenever it is of the form
1
= 0
x→±∞ x
4
= 0
x→±∞ x5
x→±∞
constant
±∞
For example,
lim
lim
6
lim √
= 0.
3
x→±∞
x
Limits at infinity often give rise to limits of the indeterminate forms
h∞i
±∞
[∞ − ∞]
and
=
.
∞
±∞
Examples: [∞ − ∞]
a) lim (x2 − x3 )
x→∞
lim (x2 − x3 ) =
x→∞
lim x3 (1/x − 1)
x→∞
= −∞
b) lim (2x3 + 4x2 − 3x + 1)
x→−∞
lim (2x3 + 4x2 − 3x + 1) =
x→−∞
lim x3 (2 + 4/x − 3/x2 + 1/x3 )
x→−∞
= −∞
Examples: [∞/∞]
5x2 − 4x + 3
x→∞ 8x2 − 7x + 9
a) lim
5x2 − 4x + 3
=
x→∞ 8x2 − 7x + 9
lim
x2 (5 − 4/x + 3/x2 )
x→∞ x2 (8 − 7/x + 9/x2 )
lim
=
5 − 4/x + 3/x2
x→∞ 8 − 7/x + 9/x2
=
5−0+0
8−0+0
=
5
8
lim
4x3 + 5x − 9
x→−∞ 7x4 − 9x2 − 6
b) lim
4x3 + 5x − 9
=
x→−∞ 7x4 − 9x2 − 6
lim
=
x3 (4 + 5/x2 − 9/x3 )
x→−∞ x4 (7 − 9/x2 − 6/x4 )
lim
4 + 5/x2 − 9/x3
x→−∞ x(7 − 9/x2 − 6/x4 )
lim
= 0
5x3 − 9
x→−∞ 8x2 + 3x − 7
c) lim
5x3 − 9
=
x→−∞ 8x2 + 3x − 7
lim
=
x3 (5 − 9/x3 )
x→−∞ x2 (8 + 3/x − 7/x2 )
lim
x(5 − 9/x3 )
x→−∞ 8 + 3/x − 7/x2
lim
= −∞
In the next two examples, it will be made use of the fact that
√
(
x2 = |x| =
x
if x ≥ 0
−x
if x < 0.
6x + 5
d) lim √
x→∞
9x2 + 4
6x + 5
lim √
=
x→∞
9x2 + 4
x(6 + 5/x)
lim p
x→∞
x2 (9 + 4/x2 )
=
x(6 + 5/x)
lim √ p
x→∞
x2 9 + 4/x2
=
x(6 + 5/x)
p
x→∞ x
9 + 4/x2
=
6 + 5/x
lim p
x→∞
9 + 4/x2
lim
6+0
= √
9+0
= 2
√
4x2 + 3
e) lim
x→−∞ 8x + 7
√
lim
x→−∞
4x2 + 3
=
8x + 7
p
x2 (4 + 3/x2 )
lim
x→−∞
x(8 + 7/x)
√ p
x2 4 + 3/x2
= lim
x→−∞
x(8 + 7/x)
p
−x 4 + 3/x2
= lim
x→−∞ x(8 + 7/x)
p
4 + 3/x2
= − lim
x→−∞ 8 + 7/x
√
4+0
= −
8+0
= −
1
4
Examples: [∞ − ∞] and [∞/∞]
√
2
a) lim
x + 8x − 5 − x
x→∞
√
x2 + 8x − 5 − x =
lim
x→∞
lim
√
x→∞
x2
√x2 + 8x − 5 + x
+ 8x − 5 − x √
x2 + 8x − 5 + x
=
(x2 + 8x − 5) − x2
lim p
x→∞
x2 (1 + 8/x − 5/x2 ) + x
=
8x − 5
lim √ p
x→∞
x2 1 + 8/x − 5/x2 + x
=
=
lim
x→∞
x(8 − 5/x)
p
x 1 + 8/x − 5/x2 + x
8 − 5/x
lim p
x→∞
1 + 8/x − 5/x2 + 1
= √
8−0
1+0−0+1
= 4
b) lim
x→−∞
3x +
lim
x→−∞
√
9x2
3x +
− 4x
√
9x2 − 4x
=
lim
x→−∞
3x − √9x2 − 4x
√
√
3x + 9x2 − 4x
3x − 9x2 − 4x
=
9x2 − (9x2 − 4x)
√
x→−∞ 3x −
9x2 − 4x
=
4x
√ p
x→−∞ 3x −
x2 9 − 4/x
=
4x
p
x→−∞ 3x − (−x) 9 − 4/x
=
4
p
x→−∞ 3 +
9 − 4/x
=
4
√
x→−∞ 3 +
9−0
=
2
3
lim
lim
lim
lim
lim