Find $$\lim_{n\to\infty}\int_{0}^{\infty}\left(1+\frac{x}{n}\right)^{-n}\sin \left(\frac{x}{n}\right)dx.$$
My attempt is find a dominating function to apply the dominated convergence theorem i.e.
$f\in L_1$ such that $\left| f_n=\left(1+\dfrac{x}{n}\right)^{-n}\right|\leq f$ for all $n$.
Thanks for any suggestion.
By letting $t=x/n$ and by using integration by parts we have that $$ \begin{align}\lim_{n\to\infty}\int_{0}^{\infty}\left(1+\frac{x}{n}\right)^{-n}\sin \left(\frac{x}{n}\right)dx&= \lim_{n\to\infty}n\int_{0}^{\infty}\frac{\sin(t)}{\left(1+t\right)^{n}}dt\\ &=\lim_{n\to\infty}\left(\frac{n}{n-1}\left[\frac{\sin(t)}{\left(1+t\right)^{n-1}}\right]_0^{+\infty}+\frac{n}{n-1}\int_{0}^{\infty}\frac{\cos(t)}{\left(1+t\right)^{n-1}}dt\right) \\&=0+1\cdot \lim_{n\to\infty}\int_{0}^{\infty}\frac{\cos(t)}{\left(1+t\right)^{n-1}}dt=0 \end{align}$$ because for $n\geq 3$, $$\left|\int_{0}^{\infty}\frac{\cos(t)}{\left(1+t\right)^{n-1}}dt\right| \leq \int_{0}^{\infty}\frac{1}{\left(1+t\right)^{n-1}}dt=\frac{1}{n-2}\to 0.$$
P.S. After integration by parts, you may also use dominated convergence.