TITLE: Limit of some integral
QUESTION [8 upvotes]: My question is how to find: 
$\displaystyle  \lim_{n \rightarrow \infty} \int\limits_0^n \frac {1}{n+n^2\sin(xn^{-2})} dx $? I've tried with a dominated convergence theorem, but it didn't work. Now, I how absolutely no idea what I can do to solve it. Please, help me.

REPLY [7 votes]: Let $t=\dfrac{x}n$. We then get
$$I_n = \int_0^1 \dfrac{dt}{1+n \sin(t/n)}$$
Now use the fact that $\sin(t/n) > \dfrac2{\pi} \cdot \dfrac{t}n$ for $t \in [0,1]$. Hence, we get that
$$1 + n \sin(t/n) > 1+\dfrac{2t}{\pi}$$
Hence,
$$I_n < \int_0^1 \dfrac{dt}{1+\dfrac{2t}{\pi}} \text{,  which is clearly bounded}$$
Hence,
$$\lim_{n \to \infty}\int_0^1 \dfrac{dt}{1+n \sin(t/n)} = \int_0^1 \lim_{n \to \infty}\dfrac{dt}{1+n \sin(t/n)} = \int_0^1 \dfrac{dt}{1+t} = \log(2)$$