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JEE Advanced 2020 Concept Recapitulation Test 4 Paper 2
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© examsnet.com
Question : 49
Total: 54
The value of
lim
t
→
0
ln
(
1
t
1
∫
0
(
1
+
2
sin
3
x
)
4
∕
x
d
x
)
is equal to:
Your Answer:
Validate
Solution:
l
=
ln
lim
t
→
0
t
∫
0
(
1
+
2
sin
3
x
)
4
∕
x
d
x
t
=
ln
lim
t
→
0
(
1
+
2
sin
3
t
)
4
∕
t
(Using L' Hospital's rule)
=
ln
e
lim
t
→
0
t
(
2
sin
3
t
)
=
lim
t
→
0
2.3
⋅
4
⋅
sin
3
t
3
t
=
24
© examsnet.com
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