A) 1
B) 0
C) - 1
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
\[{{C}_{n}}\,=\int\limits_{\frac{1}{n+1}}^{\frac{1}{n}}{\frac{{{\tan }^{-1}}(nx)}{{{\sin }^{-1}}(nx)}}\,dx;\,\,\,\,\] Put \[(nx)=t\] \[\Rightarrow \,{{C}_{n}}=\frac{1}{n}\,\int\limits_{\frac{n}{n+1}}^{1}{\frac{{{\tan }^{-1}}(t)}{{{\sin }^{-1}}(t)}dt}\] \[\therefore \,\,L\,=\underset{n\to \infty }{\mathop{\lim }}\,\,{{n}^{\mathbf{2}}}.\,{{C}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\,\,n.\,\int\limits_{\frac{n}{n+1}}^{1}{\frac{{{\tan }^{-1}}t}{{{\sin }^{-1}}t}dt}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\int\limits_{n}^{1}{\frac{{{\tan }^{-1}}t}{{{\sin }^{-1}}t}dt}\,}{\frac{1}{n}}\,\,\,\,\,\,\,\begin{matrix} (0\times \infty ) \\ \left( \frac{0}{0} \right) \\ \end{matrix}\,\,\,\,\,\,\,\,\,\] \[\therefore \] Applying Leibnitz rule and L? Hospital rule, we get limit \[=\frac{1}{2}\]
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