# Calculate $\lim_{n \rightarrow \infty} \int_{0}^{\infty} (1+\frac{x}{n})^{-n}\sin(\frac{x}{n}) dx$

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.

• The limit of the integrand is just zero for each $x$, because of the sine term. – T. Bongers Feb 21 at 18:35
• You have an inverse exponential multiplied by something going to zero. Everything is super small; DCT should allow you to interchange – Brevan Ellefsen Feb 21 at 19:37

## 1 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,

$$\lim_{n \to \infty} \int_{(0,\infty)} \left(1+ \frac{x}{n} \right)^{-n} \sin \left(\frac{x}{n}\right) \, dx = 0.$$