Limits and Continuity 61have 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^2x→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 5.2-6TIP • 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.