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

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

108 STEP 4. Review the Knowledge You Need to Score High

23. Substitutingx=0 into
    | 3 x− 4 |
    x− 2
    , you

obtain

4

− 2

=− 2.

24. Rewrite limx→ 0
    tanx
    x
    as limx→ 0
    sinx/cosx
    x

,

which is equivalent to limx→ 0

sinx

xcosx

, which

is equal to

limx→ 0

sinx

x

.limx→ 0

1

cosx

=(1)(1)= 1.

25. To find horizontal asymptotes, examine
    the limx→∞f(x) and the limx→−∞f(x). The

xlim→∞f(x)=xlim→∞

x

√

x^2 + 4

. Dividing by
the highest power ofx(and in this case, it
isx), you obtain limx→∞
x/x
√
x^2 + 4 /x

.As

x→∞,x=

√

x^2. Thus, you have

xlim→∞

√ x/x

x^2 + 4 /

√

x^2

=xlim→∞

√^1

x^2 + 4

x^2

=xlim→∞

1

√

1 +

4

x^2

=1. Thus, the liney= 1

is a horizontal asymptote.

The limx→−∞f(x)=xlim→−∞

√x

x^2 + 4

.

Asx→−∞,x=−

√

x^2. Thus, limx→−∞

√x

x^2 + 4

=x→lim−∞

√ x/x

x^2 + 4 /(−

√

x^2 )

=xlim→−∞

1

−

√

1 +

4

x^2

=−1.

Therefore, the liney=−1 is a horizontal

asymptote. As for vertical asymptotes,

f(x) is continuous and defined for all real

numbers. Thus, there is no vertical

asymptote.