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!
On
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, $(1+x/2)^{-2}$.
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.