I need to calculate the limit
$$\lim_{n \rightarrow \infty} \int_{0}^{\infty} \left(1+\frac{x}{n}\right)^{-n}\sin\left(\frac{x}{n}\right) dx$$
I tried using the dominated convergence theorem, but I couldn't find the limit of what is inside the integral, so no idea what should I do. Also, Wolfram can't calculate this integral.
Best Answer
By Bernoulli's inequality, we have
$$\left( 1+ \frac{x}{n} \right)^{n/2} \geq 1+ \frac{x}{2}$$
and therefore
$$\left( 1+ \frac{x}{n} \right)^{-n} \leq \frac{1}{(1+x/2)^2}$$ for all $x \geq 0$ and $n \in \mathbb{N}$. This implies that
$$\left|\sin \left(\frac{x}{n}\right) \right| \left| 1+ \frac{x}{n} \right|^{-n} \leq \frac{1}{(1+x/2)^2}$$
and
$$\lim_{n \to \infty} \left|\sin \left( \frac{x}{n}\right) \right| \cdot \left| 1+ \frac{x}{n} \right|^{-n} =0.$$
Hence, by the dominated convergence theorem,
$$\lim_{n \to \infty} \int_{(0,\infty)} \left(1+ \frac{x}{n} \right)^{-n} \sin \left(\frac{x}{n}\right) \, dx = 0.$$