doubt in Dirac Delta function 
Show that $D(x)=\lim_{m \to \infty}\frac{\sin(mx)}{\pi x}$ is a representation of  $\delta(x)$ in the sense of distributions, that is
$$\lim_{m \to \infty}\int_{-\infty}^{\infty} dxf(x)\frac{\sin mx}{\pi x}=f(0)$$for any smooth function $f$ of compact support. Here, $\delta(x)$ is the Dirac delta distribution: $(\delta,f)=f(0)$ for all such $f$.

 A: Here is a solution that uses known facts about Calculus (Taylor series, integration by parts and Riemann integration, improper integrals) and nothing else.

We show that for any $\phi\in\mathcal{D}(\mathbb{R})$ (smooth functions of compact support),
$$u_t(\phi)=\int_{\mathbb{R}}\frac{\sin(tx)}{x}\phi(x)\,dx\xrightarrow{t\rightarrow\infty}\phi(0)\pi$$
Suppose $\operatorname{supp}(\phi)\subset [-A,A]$. Then
\begin{aligned}
u_t(\phi)=\int^A_{-A}\frac{\sin tx}{x}\phi(x)=\int^A_{-A}\frac{\sin tx}{x}\phi(-x)dx
\end{aligned}
and so,
$$
u_t(\phi)=\int^A_{-A}\frac{\sin tx}{x}\phi_e(x)\,dx
$$
where $\phi_e(x)=\frac12(\phi(x)+\phi(-x))$ is the even part of $\phi$. The advantage of working with $\phi_e$ is that not only is $\phi_e\in\mathcal{D}(\mathbb{R})$, but also $\phi_e(0)=\phi(0)$ and $\phi'_e(0)=0$. By Taylor's theorem

*

*$\phi_e(x)=\phi(0)+O(x^2)$ around $x=0$.

*$\phi'_e(x)=O(x)$ around $x=0$.

With this in mind, we have that
$$
u_t(\phi)=\phi(0)\int^A_{-A}\frac{\sin xt}{x}\,dx +\int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx
$$
By (1) and (2), the  map $\psi(x)=\frac{\phi_e(x)-\phi(0)}{x}$, $x\neq0$ and $\psi(0)=0$,  is continuously differentiable. Integrating by parts we obtain
$$
\int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx=\frac1t\int^A_{-A}\cos(xt)\Big(\frac{\phi'_e(x)}{x} -\frac{\phi_e(x)-\phi(0)}{x^2}\Big)\,dx
$$
As $\phi'_e(x)/x$ and $(\phi_e(x)-\phi(0))/x^2$ are integrable ( Riemann integrable and thus, Lebesgue integrable) over $[-A,A]$,
$$
\Big|\int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx\Big|\leq\frac{1}{t}\left(\int^A_{-A}\Big|\frac{\phi'_e(x)}{x}\Big|+\Big|\frac{(\phi_e(x)-\phi(0)}{x^2}\Big|\,dx\right)\xrightarrow{t\rightarrow\infty}0
$$
Putting this together, we obtain that $\lim_{t\rightarrow\infty}u_t(\phi)$ exists and
$$\lim_{t\rightarrow\infty}u_t(\phi)=\lim_{t\rightarrow\infty}\phi(0)\int^A_{-A}\frac{\sin xt}{x}\,dx=\lim_{t\rightarrow\infty}\phi(0)\int^{tA}_{-tA}\frac{\sin x}{x}\,dx=\phi(0)\pi$$
That is, $u_t\xrightarrow{t\rightarrow\infty}\pi\delta_0$ in distribution.

Edit: an even nicer and (shorter if certain facts outlined below are know to the OP) has been presented by @reuns in the comment section:
define $h(x)=\int^x_{-\infty}\frac{\sin x}{x}\,dx$ (as an improper integral)
Fact: It is well known that $h(x)$ is uniformly bounded in $\mathbb{R}$ and that $\lim_{x\rightarrow\infty}h(x)=\pi$ (Not difficult to prove as a matter of fact) and so, $$\lim_{m\rightarrow\infty}h(xm)=\pi\mathbb{1}_{(0,\infty)}(x) +\frac{\pi}{2}\mathbb{1}_{\{0\}}(x)$$
With this fact in mind, consider a smooth function $f$ with compact support. Then, integrating by parts
\begin{align}
\int^\infty_{-\infty}f(x)m h'(mx)\,dx&=\int^\infty_{-\infty}f(x)\big(h(mx)\big)'\,dx = f(x)h(mx)|^\infty_{-\infty}-\int^\infty_{-\infty}h(mx)f'(x)\,dx\\
&=-\int^\infty_{-\infty}h(mx)f'(x)\,dx\xrightarrow{m\rightarrow\infty}-\pi\int^\infty_0 f'(x)\,dx\\
&=-\pi(f(\infty)-f(0))=\pi f(0)
\end{align}
the the last limit follows from dominated convergence.
As @reuns pointed out, the also holds if $f$ and $f'$ are both integrable, for then it must be that $f(\pm\infty)=0$ necessarily.

