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Question

Let Cn=1n1n+1tan1(nx)sin1(nx)dx. Then limnn2.Cn equals

A
1
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B
0
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C
1
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D
12
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Solution

The correct option is D 12
Cn=1n1n+1tan1(nx)sin1(nx)dx
Put nx=tdx=dtn
Cn=1n1nn+1tan1tsin1tdt

Now, L=limnn2Cn
=limnn1nn+1tan1tsin1tdt (×0) form
L=limn1nn+1tan1tsin1tdt1/n (00) form
Applying Leibnitz rule,
L=limn0tan1nn+1sin1nn+1(1(n+1)2)1n2
=π4×2π=12

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