I am trying to show that $$\lim_{n \rightarrow \infty} \int^{n^2}_{0}{e^{-x^2}n \sin\frac{x}{n}dx} = \frac{1}{2}.$$
I have tried by using the monotone convergence theorem, but if I take $f_n = e^{-x^2}n \sin\frac{x}{n}$ on $(0,n^2)$ and $0$ otherwise I can show neither of
1) $f_n \leq f_{n+1}$ a.e
2) $\sup_n \int f_n <\infty$
Is this the right approach or am I missing something?
We can easily prove the result using the dominated convergence theorem: let $$f_n(x)=e^{-x^2}n\sin\left(\frac xn\right)\chi_{[0,n^2]}(x)$$ then $$f_n(x)\xrightarrow{n\to\infty}xe^{-x^2}\chi_{[0,\infty)}(x)$$ and $$|f_n(x)|\le xe^{-x^2}\chi_{[0,\infty)}(x)=\varphi(x)$$ since
$$|\sin(x)|\le x,\quad \forall x\ge0$$ hence $$\lim_{n\to\infty}\int_0^\infty f_n(x)dx=\int_0^\infty xe^{-x^2}dx=-\frac12e^{-x^2}\bigg|_0^\infty=\frac12$$