Compute the following integral: $\lim_{n\rightarrow\infty}\int_0^{\infty}(1+\frac{x}{n})^{-n}\sin(\frac{x}{n})dx$ Question: Compute: $\lim_{n\rightarrow\infty}\int_0^{\infty}(1+\frac{x}{n})^{-n}\sin(\frac{x}{n})dx$. This is from Folland's Real Analysis book.
If we can find an integrable majorant, then the integral will equal $0$, since $\lim_{n\rightarrow\infty}\sin(x/n)=0$.  We can use the Binominal Theorem to show $(1+\frac{x}{n})^n\geq 1+x+(\frac{1}{2}-\frac{1}{2n})x^2$, by noticing that all the terms will be positive since $x$ only takes nonnegative values, so we just "cut it off" after those couple of terms.  Next, since $|\sin x|\leq 1$, we can say $|\frac{\sin(\frac{x}{n})}{(1+\frac{x}{n})^n}|\leq \frac{1}{1+x+(\frac{1}{2}-\frac{1}{2n})x^2}$, and $\int_0^\infty\frac{1}{1+x+(\frac{1}{2}-\frac{1}{2n})x^2}<\infty$, so we have our integrable majorant, and by Dominated Convergence Theorem we can pass the limit inside the integral giving us a value of $0$.
I am curious if there are any (unseen by me) holes in this argument, or, maybe, are there any other ways of using DCT to solve this?  Thank you!
EDIT:
Just a quick note, I wanted to use Bernouli's inequality to get $\frac{1}{(1+\frac{x}{n})^n}\leq\frac{1}{1+n\frac{x}{n}}\leq\frac{1}{1+x}$... but $\int_0^\infty\frac{1}{1+x}dx$ doesn't converge.

Let
$$I_1(n) := \int_0^n (1 + x/n)^{-n} \sin \frac{x}{n} \mathrm{d} x$$
and
$$I_2(n) := \int_n^\infty (1 + x/n)^{-n} \sin \frac{x}{n} \mathrm{d} x.$$
Clearly, $I_1(n) \ge 0$.
Using $\sin u \le u$ for all $u\ge 0$, we have
$$I_1(n) \le \int_0^n  (1 + x/n)^{-n} \frac{x}{n} \mathrm{d} x = \frac{n(1 - n 2^{1 - n})}{(n - 1)(n - 2)}.$$
(Note: $\int (1 + x/n)^r \mathrm{d} x = \frac{n}{r + 1}(1 + x/n)^{r + 1} + C$ for $r \ne -1$.)
Thus, we have $\lim_{n\to \infty} I_1(n)  = 0$.
Also, we have
$$|I_2(n)| \le \int_n^\infty (1 + x/n)^{-n} \left|\sin \frac{x}{n}\right| \mathrm{d} x
\le \int_n^\infty (1 + x/n)^{-n} \mathrm{d} x = \frac{n 2^{1 - n}}{n - 1}.$$
Thus, we have $\lim_{n\to \infty} I_2(n)  = 0$.
The desired result follows.
