Calculate limit of Lebesgue integrals

Question

I am trying to calculate this limit: $$\lim_n \int_0^{n^2}e^{-x^2}\sin(\frac{x}{n})dx$$

Since $$\int_0^{n^2}e^{-x^2}\sin(\frac{x}{n})dx=\int_{[0,\infty)}e^{-x^2}\sin(\frac{x}{n})\mathcal X_{[0,n^2]}dx,$$ then the limit of the problem is equal to $$\lim_n\int_{[0,\infty)}e^{-x^2}\sin(\frac{x}{n})\mathcal X_{[0,n^2]}dx$$

If I'm not mistaken, for each $x$, $\lim_n e^{-x^2}\sin(\frac{x}{n})\mathcal X_{[0,n^2]}=0$, so, if $f_n=e^{-x^2}\sin(\frac{x}{n})\mathcal X_{[0,n^2]}$ and I could find an integrable function $g$ such that $|f_n| \leq g$, then I could apply the dominated convergence theorem to say $$\lim_n \int_{[0,\infty)}e^{-x^2}\sin(\frac{x}{n})\mathcal X_{[0,n^2]}dx=\int_{[0,\infty)} \lim_ne^{-x^2}\sin(\frac{x}{n})\mathcal X_{[0,n^2]}dx=0$$

I couldn't do this, any suggestions to solve the problem would be appreciated.

Answer 1

These integrands are all dominated by the $\mathcal{L}^1$ function $x\mapsto e^{-x^2}$. Since they converge pointwise to zero, the integrals go there too.

Answer 2

Isn't it trivial?

$$\left|\int_{0}^{n^2}e^{-x^2}\sin\frac{x}{n}\,dx\right|\leq \frac{1}{n}\int_{0}^{n^2}x e^{-x^2}\,dx\leq \frac{1}{n}\int_{0}^{+\infty}x e^{-x^2}=\frac{1}{2n}. $$