Question

Let $C_n=\underset{\frac{1}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}(nx)}{{\sin }^{-1}(nx)}dx$ then ${\mathrm{Lim}}_{n\to \infty }{n}^2\cdot C_n$ equals

• A. 1
• B. 0
• C. -1
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$C_n=\underset{\frac{1}{n+1}}{\overset{\frac{1}{n}}{\int }}\frac{{\tan }^{-1}(nx)}{{\sin }^{-1}(nx)}dx$ (put nx $=t)\Rightarrow C_n=\frac{1}{n}\underset{\frac{n}{n+1}}{\overset{1}{\int }}\frac{{\tan }^{-1}(t)}{{\sin }^{-1}(t)}dt$
$L=\underset{n\to \infty }{\mathrm{Lim}}{n}^2\cdot C_n=\underset{n\to \infty }{\mathrm{Lim}}n\cdot \underset{\frac{n}{n+1}}{\overset{1}{\int }}\frac{{\tan }^{-1}t}{{\sin }^{-1}t}dt(\infty \times 0);L=\frac{\frac{{\tan }^{-1}t}{{\sin }^{-1}t}dt}{\frac{1}{n}}\left(\frac{0}{0}\right)$
applying Leibnitz rule
$L=\underset{n\to \infty }{\mathrm{Lim}}\frac{0-\frac{{\tan }^{-1}\frac{n}{n+1}}{{\sin }^{-1}\frac{n}{n+1}}\left(\frac{1}{(n+1{)}^2}\right)}{-\frac{1}{n^2}}=\frac{\pi }{4}\cdot \frac{2}{\pi }=\frac{1}{2}$