56 STEP 4. Review the Knowledge You Need to Score High
[–3,3] by [–3,3]
Figure 5.1-8
Example 3
Find the limit if it exists: limy→ 0
y^2
1 −cosy
.
Substituting 0 in the expression would lead to 0 /0. Multiplying both
the numerator and denominator by the conjugate (1 + cosy) produces
limy→ 0
y^2
1 −cosy
·
(1+cosy)
(1+cosy)
= limy→ 0
y^2 (1+cosy)
1 −cos^2 y
= limy→ 0
y^2 (1+cosy)
sin^2 y
= limy→ 0
y^2
sin^2 y
·
limy→ 0 (1+cos^2 y)=limy→ 0
(
y
siny
) 2
·limy→ 0 (1+cos^2 y)=
(
limy→ 0
y
siny
) 2
·limy→ 0 (1+cos^2 y)=
(1)^2 (1+1)=2. (Note that limy→ 0
y
siny
=limy→ 0
1
siny
y
=
limy→ 0 (1)
limy→ 0
siny
y
=
1
1
=1). Verify your result
with a calculator. (See Figure 5.1-9.)
[–8,8] by [–2,10]
Figure 5.1-9
Example 4
Find the limit if it exists: limx→ 0
3 x
cosx
.
Using the quotient rule for limits, you have limx→ 0
3 x
cosx
=
xlim→ 0 (3x)
xlim→ 0 (cosx)
=
0
1
=0. Verify your
result with a calculator. (See Figure 5.1-10.)
[–10,10] by [–30,30]
Figure 5.1-10