If $\phi \in C^1_c(\mathbb R)$ then $ \lim_n \int_\mathbb R \frac{\sin(nx)}{x}\phi(x)\,dx = \pi\phi(0)$.

Note that $$ \small \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\pi\phi(0)= \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\phi(0)\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}dx= \int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx $$ Denote $$ \small I_{m}(n):=\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\qquad I(n):=\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx $$ We claim that $I_m(n)$ converges to $I(n)$ uniformly by $n\in\mathbb{N}$ when $m\to\infty$. Indeed, since $\phi$ is compactly supported $$ \small \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|I_m(n)-I(n)\right|= \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\right|= $$ $$ \small \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}(-\phi(0))dx\right|= \lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\phi(0)\int\limits_{\mathbb{R}\setminus[-\pi mn,\pi mn]}\frac{\sin(y)}{y}dy\right|= $$ $$ \small |\phi(0)|\lim\limits_{m\to\infty}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(y)}{y}dy\right|=0 $$ Since convergence is uniform by $n\in\mathbb{N}$ we can say $$ \small \lim\limits_{n\to\infty} I(n)= \lim\limits_{n\to\infty} \lim\limits_{m\to\infty} I_m(n)= \lim\limits_{m\to\infty}\lim\limits_{n\to\infty} I_m(n)= \lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx $$ Since $\varphi\in C_c^1(\mathbb{R})$, then the function $x^{-1}(\varphi(x)-\varphi(0))$ is in $L^1([-\pi m,\pi m])$ for all $m\in\mathbb{N}$. Then by Riemann–Lebesgue lemma $$ \small \lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx=0 $$ so $\small\lim\limits_{n\to\infty}I(n)=0$. This is exactly what we wanted to prove.

A quick proof of this can be given using the Riemann-Lebesgue lemma, which is covered in Rudin and a number of other texts. Write your limit as $$\lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}\, dx + \lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(0)\chi_{[-1,1]}(x)}{x}\, dx $$ Here $\chi_{[-1,1]}(x)$ denotes the characteristic function of $[-1,1]$. Since $\phi(x) \in C_c^1({\mathbb R})$, the function ${\displaystyle \frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}}$ is a bounded function with compact support; the only place you have to worry about is $x = 0$ and you can use the mean value theorem for example to show it's bounded near $x = 0$. Since the function is bounded function with compact support it is in $L^1$, which is enough to apply the Riemann-Lebesgue lemma and say the first term goes to zero. As for the second term, after changing variables to $y = nx$ we may rewrite it as $$\lim_{n \rightarrow \infty} \int_\mathbb R \sin(y)\frac{\phi(0)\chi_{[-n,n]}(y)}{y}\, dy $$ $$= \phi(0)\lim_{n \rightarrow \infty} \int_{-n}^n \frac{\sin(y)}{y}\, dy $$ $$= \phi(0)\int_\mathbb R \frac{\sin(y)}{y}\, dy $$ $$= \pi \phi(0)$$ So this will be the overall limit.