Evaluate: $\lim_{n\to\infty} \int_{[0,\infty)} (1+x/n)^{-n}\sin(x/n)\,dx.$ 
Evaluate:$$\lim_{n\to\infty} \int_0^{\infty} \left(1+\frac xn\right)^{-n}\sin \frac xn\,dx.$$

I have tried to use the (Lebesgue) dominated convergence theorem to evaluate the same. At first I noticed that:
$\left|\left(1+\frac xn\right)^{-n}\sin \frac xn\right|\le\left|\left(1+\frac xn\right)^{-n}\right|\le1\text{ for all positive }x \text{ and for all natural }n\tag*{}.$
Since $g(x)=1$ is Lebesgue integrable for each $x\in[0,\infty).$ So by DCT, the given limit equals:
$$\int_0^{\infty} \lim_{n\to\infty}  \left(1+\frac xn\right)^{-n}\sin \frac xn\,dx=\int_0^{\infty}e^{-x}\cdot0\,dx=0.$$
This question was asked in our end term exam this semester and so I don't know the correct answer to this question. Therefore, just to check whether I have evaluated the limit correctly I am posting the same here on MSE. If I have gone wrong somewhere, please point out and give some insights. Thanks in advance.

Hint: Using
$$ (1+z)^n\ge 1+\binom{n}{2}z^2 $$
for $z>0$, one has
$$ (1+\frac{x}{n})^n\ge\frac{n(n-1)}{2}\frac{x^2}{n^2}=1+\frac12(1-\frac1n)x^2\ge1+\frac14x^2$$
for $n\ge2$. Then define
$$ f_n(x)=(1+\frac{x}{n})^{-n}\sin(\frac{x}{n}) $$
to estimate $f_n$ as
$$ |f_n(x)|\le\frac{4}{4+x^2} $$
for $x\ge1$ and then one can use LDC.
