Evaluate the limit $\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$ Evaluate the limit$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$$
I used L'Hopital's Rule
$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx=\lim_{n\to\infty}\frac{\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx}{\frac{1}{\sqrt{n}}}=\lim_{n\to\infty}\frac{e^{-1}\frac{1/n}{\sin (1/n)}.\frac{-1}{n^2}}{-\frac{1}{2n\sqrt{n}}}=0$$
Is it correct
 A: You did not differentiate the numerator correctly. For  a simpler proof use the inequality $\sin x \geq \frac 2 {\pi} x$ for $0 <x <\frac {\pi} 2$. You will see very easily that the limit is $0$.
[ $\int_0^{1/n}e^{-ny}dy=\frac {1-e^{-1}} n$].
A: We have
$$0 \le \sqrt{n}\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx \le \sqrt{n}\int_{0}^{\frac{1}{n}}\frac{x}{\sin x}dx.$$
By the mean value theorem there is $t_n \in (0,1/n)$ such that
$$\int_{0}^{\frac{1}{n}}\frac{x}{\sin x}= \frac{1}{n} \cdot \frac{t_n}{\sin t_n}.$$
Hence
$$0 \le \sqrt{n}\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx \le \sqrt{n}\int_{0}^{\frac{1}{n}}\frac{x}{\sin x}dx \le \frac{1}{\sqrt{n}}\cdot \frac{t_n}{\sin t_n}$$
This gives that the limit in question $=0.$
A: For $n\geq 1$,
\begin{align*}
0 < \sqrt n \int_0^{1/n} {e^{ - nx} \frac{x}{{\sin x}}dx} & < \frac{1}{{\sin 1}}\sqrt n \int_0^{1/n} {e^{ - nx} dx} \\ & < \frac{1}{{\sin 1}}\sqrt n \int_0^{ + \infty } {e^{ - nx} dx}  = \frac{1}{{\sin 1}}\frac{1}{{\sqrt n }}.
\end{align*}
Thus, by the squeeze theorem, the limit is $0$. A bit more interesting problem is
$$
\mathop {\lim }\limits_{n \to  + \infty } n\int_0^{1/n} {e^{ - nx} \frac{x}{{\sin x}}dx} .
$$
