Evaluate the limit $\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$

Question

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

Answer 1

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 $0$.

[ $\int_0^{1/n}e^{-ny}dy=\frac {1-e^{-1}} n$].

Answer 2

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 $=0.$

Answer 3

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 $$ \mathop {\lim }\limits_{n \to + \infty } n\int_0^{1/n} {e^{ - nx} \frac{x}{{\sin x}}dx} . $$