Limit of some integral 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. 
 A: 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,
$$\lim_{n \to \infty}\int_0^1 \dfrac{dt}{1+n \sin(t/n)} = \int_0^1 \lim_{n \to \infty}\dfrac{dt}{1+n \sin(t/n)} = \int_0^1 \dfrac{dt}{1+t} = \log(2)$$
A: If you change the variable with $u=x/n^2$, you get
$$
I_n=\int_0^{1/n}\frac{du}{\sin u+\frac{1}{n}}.
$$
Now pick $\epsilon>0$.
For $n$ large enough
$$
(1-\epsilon)u\leq \sin u\leq u
$$
for all $u$ in $[0,1/n]$.
So 
$$
\log 2=\int_0^{1/n}\frac{du}{u+\frac{1}{n}}\leq I_n\leq \int_0^{1/n}\frac{du}{(1-\epsilon)u+\frac{1}{n}}=\frac{\log(2-\epsilon)}{1-\epsilon}
$$
for all $n$ large enough.
So 
$$
\log 2\leq \liminf I_n\leq\limsup I_n\leq \frac{\log(2-\epsilon)}{1-\epsilon}.
$$
Letting $\epsilon$ tend to $0$, we get
$$
\lim I_n=\log 2.
$$
A: Did you try the substitution $y = {x \over n}$? That looks a lot easier to evaluate to me.
