Part II: The Answers
198
Now multiply by x
3
:
2
3
4
3
3
2
3
x
x
x
x
≤
−
( )





 ≤
sin π
Note that x
3
0
> for values of x greater than 0, so you don’t have to flip the inequalities.
Because the limit is approaching 0 from the right (but isn’t equal to 0), you can apply
the squeeze theorem to get
lim
lim
sin
lim
lim
x
x
x
x
x
x
x
x
x
→
→
→
→
+
+
+
+
≤
−
( )





 ≤
≤
0
3
0
3
2
0
3
0
3
2
3
4
0
π
3 3
0
2
−
( )





 ≤
sin π
x
Therefore, you can conclude that lim
sin
x
x
x
→
+
−
( )





 =
0
3
2
3
0
π
.
199.
5
To use lim sin
x
x
x
→
=
0
1, you need the denominator of the function to match the argument
of the sine. Begin by multiplying the numerator and denominator by 5. Then simplify:
lim
sin( ) lim
sin( )
lim
sin( )
( )
x
x
x
x
x
x
x
x
x
→
→
→
=
=
=
=
0
0
0
5
5
5
5
5
5
5
5 1
5
200.
0
Factor the 2 from the numerator and then multiply the numerator and denominator
by 1
x
. Then simplify:
lim cos
sin
lim
(cos
)
sin
lim
cos
sin
x
x
x
x
x
x
x
x
x
x
x
→
→
→
− =
−
=
−
=
0
0
0
2
2
2
1
2
1
2 2 0
1
0
( )
=
Answers
101–200