If $f$ is continuously differentiable periodic, then $n\int_{0}^{1} f(x) \sin (2\pi nx) \mathrm dx \to 0 $ If $f: \mathbb R \to \mathbb R$ is continuously differentiable periodic function of period $1$, then 
$$n\int_{0}^{1} f(x) \sin(2\pi nx)\mathrm dx \to 0 $$
as $ n\to\infty$. 

Hint. You may integrate by parts:
$$
\begin{align}
&n\int_{0}^{1} f(x) \sin(2\pi n x)\: dx\\\\
&=\left. -f(x)\frac{\cos(2\pi n x)}{2\pi}\right|_0^1+\frac1{2\pi}\int_0^1f'(x)\cos(2\pi n x)\:dx\\\\
&=\frac{f(0)-f(1)}{2\pi}+\frac1{2\pi}\int_0^1f'(x)\cos(2\pi n x)\:dx\\\\
&=0+\frac1{2\pi}\int_0^1f'(x)\cos(2\pi n x)\:dx \to 0
\end{align}
$$ as $n \to \infty$, using the Riemann-Lebesgue lemma.
