How to prove that $\lim_{k\to+\infty}\frac{\sin(kx)}{\pi x}=\delta(x)$? It is well-known that:
$$\lim_{k\to+\infty}\frac{\sin(kx)}{\pi x}=\delta(x).$$
This can also be written as
$$ 2\pi\delta(x)=\int^{+\infty}_{-\infty}e^{ikx}\,\mathrm dk.$$
However, I don't know how to prove this without using Fourier Transform. I have already searched google and looked for some books, but I just get nothing.

In short, I want to know the proof of this equation:
$$\lim_{k\to+\infty}\int^{+\infty}_{-\infty}\frac{\sin(kx)}{\pi x}f(x)dx=f(0).$$

 A: $f_k(x)=\frac{\sin(k x)}{\pi x}$ is a function with a unit integral over the whole real line. It is an entire function, its value at $x=0$ equals $\frac{k}{\pi}$ and the integral over the whole real line is concentrated in a neighbourhood of the origin that shrinks to zero as $k\to +\infty$:
$$ \lim_{k\to +\infty}\int_{-\frac{\pi}{2k}}^{\frac{\pi}{2k}}\frac{\sin(kx)}{\pi x}\,dx =\lim_{k\to +\infty}\int_{-\frac{\pi}{2k}}^{\frac{\pi}{2k}}\frac{k}{\pi}\,dx=1$$
hence:
$$ \lim_{k\to +\infty}\int_{|x|\geq\frac{\pi}{2k}}\frac{\sin(kx)}{\pi x}\,dx = 0$$
by the regularity (analiticity) of the $\text{sinc}(kx)$ in a neighbourhood of the origin. It can be seen as a consequence of the Paley-Wiener theorem, too, since the Fourier transform of $\text{sinc}(x)$ is compact-supported.
A: 
Intuitively, as $k\to\infty$, the function oscillates faster and faster, and so it "locally averages out to zero". We note that for all $a<b$:
$$\lim_{k\to\infty}\int_a^b \sin(kx) dx = \lim_{k\to\infty}\Big[\frac{\cos(kx)}{k}\Big]_a^b = 0$$
Secondly, take a test function $\varphi\in C_c^\infty(\mathbb R$). Then, for all $\epsilon>0$, we have $\frac{\varphi(x)}{\pi x}\in C^\infty_c(\mathbb R\setminus(-\epsilon,+\epsilon))$. Since this function is continuous and compactly supported, it is even uniformly continuous. In particular, it can be approximated arbitrarily well by step functions (locally constant function with finitely many discontinuities). So let $\operatorname{supp}(\varphi)\subset [-C, C]$, and take a step function $f_\delta$ such that $|\frac{\varphi(x)}{\pi x} - f_\delta(x)|\le \delta$. Then it follows
$$ \int_\epsilon^C \frac{\sin(kx)}{\pi x}\varphi(x)dx 
= \int_\epsilon^C \sin(kx) \underbrace{\Big(\frac{\varphi(x)}{\pi x} - f_\delta(x)\Big)}_{\in(-\delta, +\delta)\forall x} dx 
+ \underbrace{\int_\epsilon^C \sin(kx)f_\delta(x) dx}_{\to 0 \text{ since $f_\delta$ is step function}} $$
Hence, since $f$ can be chosen to approximate arbitrarily well, $\forall \delta>0: \lim\limits_{k\to\infty}|\int_\epsilon^C \frac{\sin(kx)}{\pi x}\varphi(x)dx| \le C\delta$; i.e. in the limit this integral must be zero. And in the neighborhood of  the origin we find:
$$ 
\int_{-\epsilon}^{+\epsilon} \frac{\sin(kx)}{\pi x}\varphi(x) dx
\sim 
\int_{-\epsilon}^{+\epsilon} \frac{\sin(kx)}{\pi x}\Big(\varphi(0)+\varphi'(0)x\Big) dx \to \varphi(0)
$$
A: I will prove that the limit is true in distribution sense: for all $\varphi \in C_{c}^{\infty}(\Bbb{R})$ we have
$$ \lim_{k\to\infty} \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x = \varphi(0). $$
The standard approximation-to-the-identity argument splits the integral into near-zero part and away-from-zero part, and estimate each part separately. In this case, however, this approach is a bit painful due to the fact that $\sin(kx)/\pi x$ is not absolutely integrable. So we adopt an indirect but easier approach.
Proof. Define $F(x) = \frac{1}{\pi}\int_{-\infty}^{x} \frac{\sin t}{t} \, \mathrm{d}t$. It is easy to check that $F$ is bounded, $F(-\infty) = 0$ and $F(+\infty) = 1$. Then performing integration by parts, we can write
$$ \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x = - \int_{\Bbb{R}} F(k x) \varphi'(x) \, \mathrm{d} x. $$
Now notice that the integrand of the RHS is uniformly bounded by $\| F \|_{\mathrm{sup}} |\varphi'| $, which is integrable. Thus taking $k \to \infty$, the dominated convergence theorem shows that
$$ \lim_{k\to\infty} \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x
= - \int_{\Bbb{R}} \lim_{k\to\infty} F(k x) \varphi'(x) \, \mathrm{d} x
= - \int_{\Bbb{R}} H(x) \varphi'(x) \, \mathrm{d} x, $$
where $H(x)$ is the Heaviside step function. Now manually computing the last integral gives
$$ - \int_{\Bbb{R}} H(x) \varphi'(x) \, \mathrm{d} x
= - \int_{0}^{\infty} \varphi'(x) \, \mathrm{d} x
= \varphi(0) $$
and therefore the claim follows.
A: Hint:You just need to prove that for any $f(x)$ which is continuous at $x=a$, the following result holds:


*

*$\int_{-\infty}^{\infty} f(x)$.$\delta(x-a)dx=f(a)$.
Added-Substituting $\delta(x-a)$ using its definition, one gets
$\int_{a}^{a+\epsilon}$$f(x).\frac{1}{\epsilon}dx$$=\frac{1}{\epsilon}\int_{a}^{a+\epsilon}$$f(x)dx=\frac{1}{\epsilon}.(a+\epsilon-a).f(c)$ (where $a\le c\le a+\epsilon$)$=f(a)$ for $\epsilon→0$
