Limits and Continuity 61
have limx→−∞
2 x+ 1
√
x^2 + 3
= xlim→−∞
2 x+ 1
√x
x^2 + 3
x
. Replacing the x below
√
x^2 +3by(−
√
x^2 ),
you have limx→−∞
2 x+ 1
√
x^2 + 3
=xlim→−∞
2 x+ 1
√x
x^2 + 3
−
√
x^2
x→lim−∞
2 +
1
x
−
√
1 +
3
x^2
=
x→lim−∞(2)−x→lim−∞
1
x
−
√
x→lim−∞(1)+xlim→−∞
(
3
x^2
)=
2
− 1
=− 2.
Verify your result with a calculator. (See Figure 5.2-6.)
[–4,10] by [–4,4]
Figure 5.2-6
TIP • Remember that ln
(
1
x
)
=ln (1)−lnx=−lnxandy=e−x=
1
ex
.
Horizontal and Vertical Asymptotes
A liney=bis called a horizontal asymptote for the graph of a functionfif either limx→∞f(x)=
bor limx→−∞f(x)=b.
A linex=ais called a vertical asymptote for the graph of a function fif either limx→a+f(x)=
+∞or limx→a−f(x)=+∞.
Example 1
Find the horizontal and vertical asymptotes of the functionf(x)=
3 x+ 5
x− 2
.
To find the horizontal asymptotes, examine the limx→∞f(x) and the limx→−∞f(x).
The limx→∞f(x)=xlim→∞
3 x+ 5
x− 2
=xlim→∞
3 +
5
x
1 −
2
x
=
3
1
=3, and the limx→−∞f(x)=xlim→−∞
3 x+ 5
x− 2
=
x→lim−∞
3 +
5
x
1 −
2
x
=
3
1
=3.
Thus,y=3 is a horizontal asymptote.