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:
$$\lim_{k\to+\infty}\int^{+\infty}_{-\infty}\frac{\sin(kx)}{\pi x}f(x)dx=f(0).$$
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.
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 $\epsilon→0$