Differential Calculus 201-NYA-05
Vincent Carrier
Asymptotes
Consider the function f whose graph is given below.
y 6
3
2
1
0
−4 −3 −2 −1
-
1
2
3
4
x
−1
−2
−3
It has a horizontal asymptote at y = −1, 1 and a vertical asymptote at x = −2, 2.
A function f has a horizontal asymptote at y = L if
lim f (x) = L
x→−∞
or
lim f (x) = L.
x→∞
It has a vertical asymptote at x = a if
lim f (x) = ±∞
x→a−
or
lim f (x) = ±∞.
x→a+
A function can have at most 2 horizontal asymptotes, but there is no upper limit on the
number of vertical asymptotes. For example, the functions tan x, sec x, csc x, and cot x
has infinitely many vertical asymptotes.
Vertical asymptotes are usually located at values of x at which a division by 0 or a
natural logarithm of 0 occurs. Thus, the behavior of f near “holes” in its domain should
be investigated.
The following functions have horizontal and/or vertical asymptote(s).
Function
Horizontal
Asymptote
Vertical
Asymptote
1/x
y=0
x=0
tan x
x = (2k + 1)π/2
k∈Z
sec x
x = (2k + 1)π/2
k∈Z
csc x
x = kπ
k∈Z
cot x
x = kπ
k∈Z
arctan x
y = ± π/2
ex
y=0
ln x
x=0
ln |x|
x=0
Examples:
a) f (x) = x3 + 6x2 + 9x
D=R
Horizontal Asymptotes:
lim (x3 + 6x2 + 9x)
lim f (x) =
x→−∞
x→−∞
lim x3 (1 + 6/x + 9/x2 )
=
x→−∞
= −∞
lim f (x) =
x→∞
lim (x3 + 6x2 + 9x)
x→∞
= ∞
Vertical Asymptotes: None.
b) f (x) =
x2
x
−9
D = R \ {−3, 3}
Horizontal Asymptotes:
lim f (x) =
x→−∞
=
lim
x→−∞ x2
x 1/x
− 9 1/x
1
x→−∞ x − 9/x
lim
= 0
lim f (x) =
x→∞
=
lim
x→∞ x2
x 1/x
− 9 1/x
1
x→∞ x − 9/x
lim
= 0
Vertical Asymptotes:
x = −3
x=3






x
lim − 2
= −∞
x→−3 x − 9
−
0+





x
lim + 2
= ∞
x→−3 x − 9
−
0−

x


lim− 2
= −∞


 x→3 x − 9




 lim
+
x→3
x
= ∞
2
x −9
+
0−
+
0+
c) f (x) = (x + 3)2 e−(x+1)
D=R
Horizontal Asymptotes:
lim f (x) =
x→−∞
lim (x + 3)2 e−(x+1)
x→−∞
= ∞
lim f (x) =
x→∞
=
H
=
H
=
lim (x + 3)2 e−(x+1)
x→∞
(x + 3)2
x→∞
ex+1
lim
2(x + 3)
x→∞
ex+1
lim
lim
= 0
Vertical Asymptotes: None.
2
x→∞ ex+1