TITLE: Integral limit of $\sin(x/n)f(x)$
QUESTION [6 upvotes]: For any $f\in L^1[0,\pi]$, evaluate
$n\to \infty \int^\pi_0 n$sin$(x/n)f(x)dx$
My idea is, $n$sin$(x/n)f(x)\to xf(x)$ and it seems that it is increasing sequence. I am not able to show it is increasing. Next thing if it increasing how could we apply monotone convergence theorem unless f is positive. 
Next idea, i tried substitution, taking $x/n$ as $t$ but i get $f(tn)$ after substitution. Then stopped there. Do you have any specific idea for this..

REPLY [5 votes]: Since 
$$\sin\left(\frac x n\right)\sim_\infty \frac xn$$
then we have
$$ \forall x\in[0,\pi],\quad n \sin\left(\frac x n\right)f(x)\xrightarrow{n\to\infty} xf(x)$$
moreover using that $|\sin x|\le |x|,\;\forall x$ we have
$$\left|n\sin\left(\frac x n\right)f(x)\right|\le |xf(x)|\le \pi|f(x)|\in L^1[0,\pi]$$
so by the dominated convergence theorem we have
$$\lim_{n\to\infty}\int_0^\pi n\sin\left(\frac x n\right)f(x)=\int_0^\pi xf(x)dx$$