Using monotone convergence theorem to find limit of integral

Question

Consider the following limit: $$\lim_{n \rightarrow \infty} \int_{0}^{\pi} \frac{5nxe^{-nx}}{1+\sin^2(3x/n)} d\lambda(x).$$ I'd like to find this limit using the Monotone Convergence Theorem. I define the sequence of functions $f_n(x)=\frac{5nxe^{-nx}}{1+\sin^2(3x/n)}$.
I have not been able to show that it is increasing, i.e. $f_n(x) \leq f_{n+1}(x)$ for all $n$ and $\lambda$-almost every $x$. I have tried looking at $\frac{f_n(x)}{f_{n+1}(x)}$ and show it is less than or equal to $1$, but I don't think it works.
Also, assuming I can show the sequence is increasing, when I change the order of the limit and integral, I obtain $$\int_{0}^{\pi} \lim_{n \rightarrow \infty} \frac{5nxe^{-nx}}{1+\sin^2(3x/n)} dx=\int_{0}^{\pi} \lim_{n \rightarrow \infty} x \frac{5n}{e^{nx}(1+\sin^2(3x/n))}dx=0.$$ Is this right? Am I using the correct approach?

Answer 1

The sequence $(f_n)$ is not increasing, but you could note that $$\frac{5nxe^{-nx}}{1+\sin^2(3x/n)}\leq 5nxe^{-nx}$$ for all $n\in\mathbb{N}$ and $x\in[0,\pi]$. Then compute $$\mathcal{I}_n=\int_0^{\pi}5nxe^{-nx}dx$$ using partial integration and you should find that $\mathcal{I}_n\to0$ as $n\to\infty$.

Answer 2

You could also let $x=y/n$ to see the integral equals

$$\frac{1}{n}\int_0^{n\pi} \frac{5ye^{-y}}{1+\sin^2(3y/n^2)}\,dy \le \frac{1}{n}\int_0^{n\pi} 5ye^{-y}\,dy<\frac{1}{n}\int_0^{\infty} 5ye^{-y}\,dy .$$

The last integral is finite, so the terms are no more than a constant times $1/n,$  hence $\to 0.$