5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book May 7, 2018 9:52

Limits and Continuity 97

To find the vertical asymptotes, look forxvalues such that the denominator (x−2) would
be 0, in this case,x=2. Then examine:

(a) limx→ 2 +f(x)=lim

x→ 2 +

3 x+ 5

x− 2

=

xlim→ 2 +(3x+5)

xlim→ 2 +(x−2)

, the limit of the numerator is 11 and the limit

of the denominator is 0 through positive values, and thus, lim

x→ 2 +

3 x+ 5

x− 2

= ∞.

(b) limx→ 2 −f(x)=limx→ 2 −
3 x+ 5
x− 2

=

limx→ 2 −(3x+5)

limx→ 2 −(x−2)

, the limit of the numerator is 11 and the limit

of the denominator is 0 through negative values, and thus, limx→ 2 −

3 x+ 5

x− 2

=−∞.

Therefore,x=2 is a vertical asymptote.

Example 2

Using your calculator, find the horizontal and vertical asymptotes of the functionf(x)=
x
x^2 − 4

.

Entery 1 =
x
x^2 − 4

. The graph shows that asx →±∞, the function approaches 0, thus

xlim→∞f(x)=x→lim−∞f(x)=^0 .Therefore, a horizontal asymptote isy=0 (or thex-axis).
For vertical asymptotes, you notice that lim
x→ 2 +
f(x) = ∞, lim
x→ 2 −
f(x) =−∞, and
lim
x→− 2 +

f(x)=∞, lim

x→− 2 −

f(x)=−∞. Thus, the vertical asymptotes arex=−2 andx=2.

(See Figure 6.2-7.)

[−8, 8] by [−4, 4]

Figure 6.2-7

Example 3

Using your calculator, find the horizontal and vertical asymptotes of the functionf(x)=
x^3 + 5
x

.

Entery 1 =
x^3 + 5
x

. The graph off(x) shows that asxincreases in the first quadrant, f(x)

goes higher and higher without bound. Asxmoves to the left in the second quadrant, f(x)
again goes higher and higher without bound. Thus, you may conclude that limx→∞f(x)=∞
and limx→−∞f(x)=∞and thus, f(x) has no horizontal asymptote. For vertical asymptotes,