Answers and Explanations 197
Answers
101–200
196.
0
Notice that for all values of x except for x = 0, you have − ≤





 ≤
1
2
1
2
cos x
due to
the range of the cosine function. So for all values of x except for x = 0, you have
− ( ) ≤ ( )





 ≤ ( )
1
2
1
4
4
2
4
x
x
x
x
cos
Because x is approaching 0 (but isn’t equal to 0), you can apply the squeeze theorem:
lim
lim cos
lim
lim cos
x
x
x
x
x
x
x
x
x
x
→
→
→
→
−
( ) ≤





 ≤
( )
≤
0
4
0
4
2
0
4
0
4
2
0
2
2 2
0





 ≤
Therefore, you can conclude that lim cos
x
x
x
→





 =
0
4
2
2
0.
197.
0
Notice that for all values of x > 0, you have − ≤





 ≤
1
2
1
sin x
due to the range of
the sine function. So for all values of x > 0, you have
− ( ) ≤





 ≤ ( )
1
2
1
2
2
2
x
x
x
x
sin
Because x is approaching 0 (but isn’t equal to 0), you can apply the squeeze theorem:
lim
lim
sin
lim
lim
si
x
x
x
x
x
x
x
x
x
→
→
→
→
+
+
+
+
−
( ) ≤





 ≤
( )
≤
0
2
0
2
0
2
0
2
2
0
n n 2
0
x





 ≤
Therefore, you can conclude that lim
sin
x
x
x
→
+





 =
0
2
2
0.
198.
0
Notice that for all values of x except for x = 0, you have − ≤
( ) ≤
1
1
2
sin π
x
due to
the range of the sine function. So for all values of x except for x = 0, you have the
following (after multiplying by –1):
− ≤ −
( ) ≤
1
1
2
sin π
x
You need to make the center expression match the given one, x
x
3
2
3 −
( )






sin π , so do
a little algebra. Adding 3 gives you
− + ≤ −
( ) ≤ +
≤ −
( ) ≤
1 3 3
1 3
2 3
4
2
2
sin
sin
π
π
x
x