Question about Dominated Convergence Theorem. 
How to compute $$\lim_{n \to \infty}\int_0^{\infty}\Big(1+\frac{x}{n}\Big)^{-n}\sin\Big(\frac{x}{n}\Big) dx$$

I want to use the Dominated Convergence Theorem. so it becomes $$\int_0^{\infty}\lim_{n\to \infty}\Big(1+\frac{x}{n}\Big)^{\frac{n}{x}(-x)}\sin\Big(\frac{x}{n}\Big) dx=\int_0^{\infty}e^{-x}\times 0 dx=0$$
I have trouble finding the dominating term, which is essential for me to apply dominated convergence theorem.
Could anyone help me with this? Thanks!
 A: By the Bernoulli inequality we have:
$$ \left(1+\frac{x}{n}\right)^{-n}\leq e^{-x}\left(1+\frac{x}{n}\right)\tag{1}$$
and:
$$\begin{eqnarray*} \int_{0}^{+\infty}e^{-x}\left(1+\frac{x}{n}\right)\sin\frac{x}{n}\,dx &=& n \int_{0}^{+\infty}(1+x)\,\sin(x)\, e^{-nx}\,dx\\ &\leq& n\int_{0}^{+\infty}(1+x)\,e^{-nx}\,dx\\&=&1+\frac{1}{n}\tag{2}\end{eqnarray*} $$
So the dominated convergence theorem applies. In fact, 
$$ \int_{0}^{+\infty}(1+x)\,\sin(x)\,e^{-nx}\,dx = \left(\frac{n+1}{n^2+1}\right)^2\leq\frac{1}{(n-1)^2}\tag{3}$$
so we don't even need the dominated convergence theorem to prove that the limit is zero.
A: The sine is clearly bounded by 1, so it suffices to show that
$$ \left( 1+\frac{x}{n} \right)^{-n} \leqslant \frac{1}{(1+x/2)^2}, $$
which is clearly integrable. How do you do that? In fact, you can do it in the same way that G.H. Hardy shows in A Course in Pure Mathematics that $(1+1/n)^n$ is increasing in $n$ (viz., that the size of each term increases, and the number of terms also increases). In a similar manner, the sequence $(1+x/n)^{-n}$ can be shown to be decreasing, and hence you immediately conclude that it is bounded by the second term, $(1+x/2)^{-2}$.
