In Arfken's "Mathematical Methods for Physicists", in Section 20.2 "Fourier Transform":
It is said that "We write the fairly obvious equation $f(x)=\lim_{n\to\infty} \int_{-\infty}^{+\infty} f(t) \delta_{n}(t-x)dt$..." where $\delta_{n}(t)=\frac{1}{2\pi} \int_{-n}^{+n} e^{i\omega t} d\omega$ (delta function).
I don't understand why $f(x)=\lim_{n\to\infty} \int_{-\infty}^{+\infty} f(t) \delta_{n}(t-x)dt$. Could someone explain?
A short, non-rigorous "physicist's proof" is as follows:
We can compute $$ \delta_n(t) = \frac{1}{\pi t}\sin(n t).$$ Then, we have $$ \lim_{n\to \infty}\int_{-\infty}^\infty f(t)\delta_n(t-x)dt = \lim_{n\to \infty}\int_{-\infty}^\infty f(t) \frac{1}{\pi (t-x)}\sin(n(t-x))dt \\ = \frac{1}{\pi}\int_{-\infty}^\infty \lim_{n\to \infty}f\left(x + \frac{u}{n}\right)\frac{\sin(u)}{u}du \\ =f(x)\frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin u}{u}du = f(x).$$