Part II: The Answers
502
Answers
801–900
lim
ln
ln
lim
ln
x
x
x
x
x x
x
x
x
x
→
→
+
+
− −
( ) −
(
)


 


  =
−
−
(
)+ ( )
1
1
1
1
1 1
1
1
1
 









Multiply the numerator and denominator by x:
lim
ln
x
x
x
x x
→
+
−
− +
1
1
1
Substituting in the value x = 1 gives you the indeterminate form 0
0
, so you can again use
L’Hôpital’s rule:
lim
ln
lim
ln
lim
ln
x
x
x
x
x
x x
x x x
x
→
→
→
+
+
+
−
− +
=
+ ( ) + ( )
=
+
+
=
1
1
1
1
1
1
1 1
1
1
1
1
1 1
1
1 1
1
2
+ ( )+
=
ln
832.
−
( )+
1
5
5
cos x C
Here’s the given integral:
sin 5x dx
( )
∫
If you let u = 5x, then du = 5dx, or 1
5
du dx
= . Substituting into the original integral,
you get
sin
s in
cos
cos
u du
u du
u
C
x C
( )
=
( )
=
−
( )
(
) +
= −
( )+
∫
∫
1
5
1
5
1
5
1
5
5
833.
1
101
4
101
x
C
+
(
) +
Here’s the given integral:
x
d x
+
(
)
∫ 4
100
If you let u = x + 4, then du = dx. Substituting into the original integral gives you the
following:
u du u
C
x
C
100
101
101
101
1
101
4
∫
=
+
=
+
(
) +