TITLE: Let $C_n =\int^{1/n}_{1/(n+1)}\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx $
QUESTION [6 upvotes]: Let
$$C_n =\int^{1/n}_{1/(n+1)}\frac{\tan^{-1}(nx)}{\sin^{-1}(nx)}dx\textrm{.} $$
Then $\lim_{n \to \infty} n ^2C_n$ equals :

(a) $1$
(b) $0$
(c) $-1$
(d) $\frac12$
I am not getting any clue on this. How to proceed? Please help on this. I will be grateful to you.

REPLY [6 votes]: You can use the Mean Value Theorem for integrals to get the limit. In fact, since
$$ \int_{\frac{1}{n+1}}^{\frac1n}\frac{\arctan nx}{\arcsin nx}dx=\frac{\arctan n\xi_n}{\arcsin n\xi_n}\frac{1}{n(n+1)}$$
where 
$$\frac{1}{n+1}<\xi_n<{\frac1n},$$
we have
\begin{eqnarray}
\lim_{n\to\infty}n^2C_n&=&\lim_{n\to\infty}n^2\int_{\frac{1}{n+1}}^{\frac1n}\frac{\arctan nx}{\arcsin nx}dx\\
&=&\lim_{n\to\infty}n^2\frac{\arctan n\xi_n}{\arcsin n\xi_n}\frac{1}{n(n+1)}\\
&=&\lim_{n\to\infty}\frac{\arctan n\xi_n}{\arcsin n\xi_n}.
\end{eqnarray}
Since 
$$\frac{1}{n+1}<\xi_n<{\frac1n}$$
we have
$$\frac{n}{n+1}<n\xi_n<1$$
or $\lim_{n\to\infty}n\xi_n=1$. So
$$ \lim_{n\to\infty}n^2C_n=\frac{\arctan 1}{\arcsin 1}=\frac{\frac{\pi}{4}}{\frac{\pi}{2}}=\frac12.$$
So you have to choose (d).