Computing $\lim_{n \to \infty} \int_0^{n^2} e^{-x^2}n\sin\frac{x}{n}\,dx$? 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!
 A: 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
$$\lim_{n \to \infty} \int_0^{n^2} e^{-x^2} n \sin \left( \frac{x}{n}\right) dx = \int_0^\infty \lim_{n\to \infty} f_n(x)dx  = \int_0^\infty x e^{-x^2} dx =\frac{1}{2}.$$
A: Notice that 
$$\int_0^{n^2} n\sin(x/n)e^{-x^2} dx=\int_0^{n^2} \left(n\sin(x/n)-x\right)e^{-x^2} dx + \frac{1-e^{-n^4}}{2}.$$
Moreover $|\sin(t)-t|\leq t^2$ implies $\left|n\sin(x/n)-x\right|\le n(x/n)^2=x^2/n$. Therefore
$$\left|\int_0^{n^2} \left(n\sin(x/n)-x\right)e^{-x^2} dx\right|\leq 
\int_0^{n^2} \left|\left(n\sin(x/n)-x\right)\right|e^{-x^2} dx\leq
\frac{1}{n}\int_0^{\infty} x^2e^{-x^2} dx$$
and the right-side goes to zero as $n\to+\infty$.
It follows that
$$\lim_{n\to \infty}\int_0^{n^2} n\sin(x/n)e^{-x^2} dx=\frac{1}{2}.$$
