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JEE Main Math Class 12 Limits and Derivatives Part 1 Questions
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© examsnet.com
Question : 96
Total: 100
lim
x
→
π
2
[
1
−
tan
(
x
2
)
]
[
1
−
sin
x
]
[
1
+
tan
(
x
2
)
]
[
π
−
2
x
]
3
is
[2003]
∞
1
8
0
1
32
Validate
Solution:
lim
x
→
π
2
tan
(
π
4
−
x
2
)
.
(
1
−
sin
x
)
(
π
−
2
x
)
3
Let
x
=
π
2
+
y
;
y
→
0
=
lim
y
→
0
tan
(
−
y
2
)
.
(
1
−
cos
y
)
(
−
2
y
)
3
=
lim
y
→
0
−
tan
y
2
2
sin
2
y
2
(
−
8
)
.
y
3
8
.
8
[
∵
1
−
cos
θ
=
2
sin
2
θ
2
]
=
lim
y
→
0
1
32
tan
y
2
(
y
2
)
.
[
sin
y
∕
2
y
∕
2
]
2
=
1
32
[
∵
lim
θ
→
0
sin
θ
θ
=
lim
θ
→
0
tan
θ
θ
=
1
]
© examsnet.com
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