limx→2π(x−2π)tan2z
please give me solution
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lt
x→
2
π
(1+tan
2
x
)(π−2x)
3
(1−tan
2
x
)(1−sinx)
=lt
x→
2
π
(π−2x)
3
tan(
4
π
−
2
x
)(1−sinx)
=lt
x→
2
π
(2x−π)
3
(sinx−1)tan(
4
π
−
2
x
)
Letx=
2
π
+h
Now,
lt
h→0
(2h)
3
−(cosh−1)tan(
2
h
)
=lt
h→0
8h
3
(1−cosh)tan(
2
h
)
=
8
1
lt
h→0
(
h
2
1−cosh
)
h
tan(
2
h
)
=lt
h→0
2
h
×2
tan(
2
h
)
=
32
1
which is the required answer.
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