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IIT JEE Advanced 2006 Paper 1
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© examsnet.com
Question : 89
Total: 120
For x > 0,
lim
x
→
0
[
(
s
i
n
x
)
1
x
+
(
1
x
)
s
i
n
x
]
is
0
- 1
1
2
Validate
Solution:
For x > 0, we have
lim
x
→
0
+
[
(
s
i
n
x
)
1
x
+
(
1
x
)
s
i
n
x
]
Therefore,
lim
h
→
0
+
[
(
s
i
n
h
)
1
h
+
(
1
h
)
s
i
n
h
]
Now,
lim
h
→
0
[
0
+
e
s
i
n
h
l
n
(
1
h
)
]
=
lim
h
→
0
e
[
s
i
n
h
l
n
(
1
h
)
]
[Since
lim
h
→
0
(
s
i
n
h
)
1
h
= 0]
=
lim
h
→
0
e
(
−
l
n
h
1
∕
s
i
n
h
)
= 1 (by using L’ Hospital’s rule)
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