Compute the value of $\lim \int_0^1 f_n(x) \sin (nx)dx.$ 
Let $\{f_n(x)\}$ be a sequence in $L^2[0,1]$ and $\lim f_n=f$ almost everywhere, where $f(x)\in L^2[0,1]$. Then find
$$\lim_{n\to \infty} \int_0^1 f_n(x)\sin (nx)dx.$$

By Riemann-Lebesgue lemma, we can conclude that $$\lim \int_0^1 f(x)\sin(nx)dx=0.$$
But how to compute it for $f_n(x)$?

The limit need not exist. For example if $f_n(x)=n^{2} \chi_{(0,\frac  1n)}$ and $f=0$ then $f_n \to f$ at every point but the given integral tends to $\infty$.
[$\int_0^{1}f_n(x) \sin (nx)dx=n\int_0^{1} \sin y dy=n(1-\cos 1)$ by the substitution $y=nx$].
