Answers and Explanations 201
205.
1
5
Begin by factoring the denominator and rewriting the limit:
lim
sin
lim
sin
lim
sin
x
x
x
x
x
x
x
x
x
x
→
→
→
−
(
)
+ −
=
−
(
)
−
(
) +
(
)
=
−
2
2
2
2
2
6
2
2
3
2
( (
)
−
(
)
⋅
+
→
x
x
x
2
1
3
2
lim
Notice that you can rewrite the first limit using the substitution θ = −
x 2 so that
as x → 2, you have θ → 0; this step isn’t necessary, but it clarifies how to use
lim sin
x
x
x
→
=
0
1 in this problem.
Replacing (x – 2) with θ and replacing x → 2 with θ → 0 in the limit lim
sin(
)
(
)
x
x
x
→
−
−
2
2
2
gives
you the following:
lim
sin
lim
lim
sin
lim
x
x
x
x
x
x
x
→
→
→
→
−
(
)
−
(
)
⋅
+
=
( )
( )
⋅
+
=
2
2
0
2
2
2
1
3
1
3
θ
θ
θ
1 1
1
2 3
1
5
( ) +
( )
=
206.
1
2
Begin by rewriting tan x and then multiply the numerator and denominator by 1
x
. Then
simplify:
lim sin
tan
lim sin
sin
cos
lim
(sin )
si
x
x
x
x
x
x
x
x
x
x
x
x
x
x
→
→
→
+
=
+
=
+
0
0
0
1
1
n n
cos
lim
sin
sin
cos
lim
sin
sin
cos
x
x
x
x
x
x
x
x
x
x
x
x
x
x
(
)
=
+
=
+
( )
→
→
0
0
1
1
1
Answers
201–300