I am trying to compute this integral/limit, I don't feel like I have any good insight...
$$\lim_{n \to \infty} \int_0^{n^2} e^{-x^2}n\sin\frac{x}{n} \, dx.$$
I have tried to make a change of variable to get rid of the $n^2$, I changed to $X=\frac{x}{n^2}$ but got something even worse, I've tried to reach a situation where I could use a convergence theorem for Lebesgue Integrals,... I'm not sure I'm even on the right track!
Could you give me a hint on how to start this?
Thank you very much!
Observe that $$\lim_{n \to \infty} \int_0^{n^2} e^{-x^2} n \sin \left( \frac{x}{n}\right) dx = \lim_{n\to \infty} \int_0^\infty f_n(x)dx $$ where $$f_n(x) = e^{-x^2} n \sin \left( \frac{x}{n}\right) \textbf{1}_{[0,n^2]}(x)$$ where $\textbf{1}$ denotes the indicator function.
Now observe that $$\lim_{n\to \infty} n \sin \left( \frac{x}{n}\right) = x$$ uniformly and $$|n \sin \left( \frac{x}{n}\right)| \leq |x|$$ for all $n\geq 1$. As a result, $$|f_n(x)| \leq |x| e^{-x^2}$$ which is clearly integrable on $(0,\infty)$. Hence we can push the limit inside the integral by Lebesgue's dominated convergence theorem: Thus $$\lim_{n \to \infty} \int_0^{n^2} e^{-x^2} n \sin \left( \frac{x}{n}\right) dx = \int_0^\infty \lim_{n\to \infty} f_n(x)dx = \int_0^\infty x e^{-x^2} dx =\frac{1}{2}.$$