
15-56 Calculus – Differentiation and Integration
=
+
∑
=
+
∫
=
→∞
=
lim
l
n
r
n
n
r
n
dx
x
1
1
1
1
1
0
1
(Step 1)
(Step 2)
oog( ) log log log1 2 1 2
0
1
+
[ ]
= − =x (Step 3).
Example 15.47 Evaluate
lim sin sin sin
n
n n n
n
n
→∞
+ + +
1 1 2
Solution:
lim sin sin sin lim sin
n n
r
n
n n n
n
n n
r
n
→∞ →∞
=
+ + +
=
∑
1 1 2 1
1
(Stepp 1)
(Step 2)
(Step 3).
=
∫
= −
[ ]
= − + = −
sin
cos cos cos
x dx
x
0
1
0
1
1 1 1 1
Example 15.48 Find the limit of
n
n
n
n
n
n
n
n n
3
3 3
3
4 8
4 1
+
+
+
+
+ +
+ −
( ) ( )
( )
.
Solution: General term
n
n r
n
r
n
+ −
[ ]
=
+
−
4 1
1 1
1
4 1
3 3
( ) ( )
Consider
( )r
n
r
n
+ =
+
1
1 1
1
4
3
th
term
Thus,
n
n
n
n
n
n
n
n n
n
r
n
n
r
n
3
3 3
3
1
4 8
4 1
1 1
1
4
+
+
+
+
+ +
+ −
=
+
∑
→∞
=
( ) ( )
( )
lim
((Step ...