If $\phi \in C^1_c(\mathbb R)$ then $ \lim_n \int_\mathbb R \frac{\sin(nx)}{x}\phi(x)\,dx = \pi\phi(0)$. 
Let $\phi \in C^1_c(\mathbb R)$. Prove that $$
\lim_{n \to +\infty} \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx = \pi\phi(0).
$$

Unfortunately, I didn't manage to give a complete proof. First of all, I fixed $\varepsilon>0$. Then there exists a $\delta >0$ s.t. 
$$
\vert x \vert < \delta \Rightarrow \vert \phi(x)-\phi(0) \vert < \frac{\varepsilon}{\pi}.
$$
Now, I would use the well-known fact that 
$$
\int_\mathbb R \frac{\sin x}{x} \, dx = \pi.
$$
On the other hand, by substitution rule, we have also
$$
\int_\mathbb R \frac{\sin(nx)}{x} \, dx = \int_\mathbb R \frac{\sin x}{x} \, dx = \pi.
$$
Indeed, I would like to estimate the quantity
$$
\begin{split}
& \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \pi \phi(0) \right\vert = \\ 
& = \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \phi(0)\int_\mathbb R \frac{\sin{(nx)}}{x}dx \right\vert \le \\ 
& \le \int_\mathbb R \left\vert \frac{\sin(nx)}{x}\right\vert \cdot \left\vert  \phi(x)-\phi(0) \right\vert dx
\end{split}
$$
but the problem is that $x \mapsto \frac{\sin(nx)}{x}$ is not absolutely integrable over $\mathbb R$. Another big problem is that I don't see how to use the hypothesis $\phi$ has compact support.
I think that I should use dominated convergence theorem, but I've never done exercises about this theorem. Would you please help me? Thank you very much indeed.

Assume that $\phi(x)$ is supported in $|x|< L$. Since $\phi$ is differentiable, $\frac{\phi(x)-\phi(0)}{x}$ is bounded and therefore integrable on $|x|<L$.
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x
&=\pi\,\phi(0)+\int_{-\infty}^\infty\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x\\
&=\pi\,\phi(0)+\color{#C00000}{\int_{-L}^L\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x}\\
&-2\,\phi(0)\color{#00A000}{\int_{nL}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x}\\
\end{align}
$$
As $n\to\infty$, the red integral vanishes by the Riemann-Lebesgue Lemma and the green integral vanishes because The Dirichlet Integral converges. This leaves us with
$$
\lim_{n\to\infty}\int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x=\pi\,\phi(0)
$$
