Answers and Explanations 489
1
9
1
9 1 9
1
3 1 3
2
0
3 4
2
0
3 4
2
0
3 4
−
=
−






=
− ( )
∫
∫
∫
u
du
u
du
u
du
Now use a new substitution w u
= 3
so that dw
du
= 1
3
, or 3
 
dw = du. You can again
find the new limits of integration by noting that if u = 3
4
, then w =
=
3 4
3
1
4
, and if
u = 0, then w = =
0
3
0. With these substitutions, you get the following:
3
3 1
1
4
0
1
4
2
0
1 4
1
0
1 4
1
1
1
dw
w
w
−
=
=
( ) −
=
( )
∫
−
−
−
−
sin
sin
s in
sin
800.
3
The given problem is
lim
x
x
x
→−
+
+
1
3
1
1
Notice that if you substitute in the value x = –1, you get the indeterminate form 0
0
.
Now use L’Hôpital’s rule to find the limit:
lim
lim
x
x
x
x
x
→−
→−
+
+
=
= −
( ) =
1
3
1
2
2
1
1
3
1
3 1
3
801.
4
3
Here’s the given problem:
lim
x
x
x
→
−
−
1
4
3
1
1
Notice that if you substitute in the value x = 1, you get the indeterminate form 0
0
.
Using L’Hôpital’s rule gives you the limit:
lim
lim
lim
x
x
x
x
x
x
x
x
→
→
→
−
−
=
=
=
( ) =
1
4
3
1
3
2
1
1
1
4
3
4
3
4 1
3
4
3
802.
1
5
Start with the given problem:
lim
x
x
x
x
→
−
+ −
2
2
2
6
Answers
801–900