5 Steps to a 5 AP Calculus BC 2019

Limits and Continuity 73

19. (See Figure 5.7-3.) Ifb=0, thenr=0,
    butrcannot be 0. Ifb=−3,−2, or− 1 f
    would have more than one root. Thus
    b=1. Choice (E).

(–2, 3)

1

• 2


• 22


• 1

(2, 4)

y

x

0

Figure 5.7-3

20. Substitutingx=0 would lead to 0/0.
    Substitute (1−cos^2 x) in place of sin^2 x
    and obtain

xlim→ 0

1 −cosx

sin^2 x

=limx→ 0

1 −cosx

(1−cos^2 x)

=limx→ 0

1 −cosx

(1−cosx)(1+cosx)

=limx→ 0

1

(1+cosx)

=

1

1 + 1

=

1

2

.

Verify your result with a calculator. (See

Figure 5.7-4)

[–10,10] by [–4,4]

Figure 5.7-4

5.8 Solutions to Cumulative Review Problems

21. Rewrite 3x− 2 y=6iny=mx+bform,
    which isy=

3

2

x− 3 .The slope of this line

whose equation isy=

3

2

x−3ism=

3

2

.

Thus, the slope of a line perpendicular to

this line ism=−

2

3

. Since the
perpendicular line passes through the
point (2,−4), therefore, an equation of
the perpendicular line is
y−(−4)=−

2

3

(x−2), which is equivalent

toy+ 4 =−

2

3

(x−2).

22. The graph indicates that limx→ 4 − f(x)=3,
    f(4)=1, and limx→ 4 f(x) does not exist.
    Therefore, only statements I and III are
    true.


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.