We are supposed to calculate the limit of the integral $\lim\_{n\to\infty}\int\_0^\pi \frac{\sin nx}{nx}dx$.

What I am currently thinking is this. Since $|\frac{\sin nx}{nx}|<\frac{1}{n}$, we can prove that this converges uniformly and can then use the dominated convergence theorem to shift the limit inside the integral, as in 
$\int\_0^\pi \lim\_{n\to\infty}\frac{\sin nx}{nx}dx = \int\_0^\pi \frac{1}{n} \,dx$.

However, I don't think that this is correct. Does someone know the correct way to solve this?