Let $\delta(x)$ be the Dirac Delta function https://en.wikipedia.org/wiki/Dirac_delta . It can be defined in the weak sense as the limit of a Gaussian
$$\delta(x)=\lim\limits_{\epsilon\to 0^+}\frac{e^{-\frac{1}{2}\frac{x^2}{\epsilon^2}}}{\epsilon\sqrt{2\pi}}$$
Could you provide a reference to a book that proves this rigorously and authoritatively, please? The statement is obvious as is, but I just need an authoritative reference. I could search for a reference myself, but lesser deviant creatures with huge brain damage and augmented instinctive impulses of a half-year old who want to bully you and steal your bubble gum and your sandwich and your canonically conjugate variables energy and time, driven by an instinctive fear of being left alone and die as the creatures they truly know they are: the scum of Earth, feeding on other people's lives, disabling them in their growth, posing as superior, charging the worker to pay the price for their very own work, go figure this, how damaged should your genetics be to do this, without actually paying them for their work, just like, say, banks, and who pose as humans as well as scientists among other normal and human scientists, keep those books behind the paywall, so finding it for me falls under the Fundamental no-go theorem of the pigsty, unfortunately... Yeah, we both live in this... And it stinks.
It also goes under the name Gauss-Weierstrass or Mellin-Gauss-Weierstrass function or distribution or kernel or transform, in one representation or another...
One use of it is, for instance, in Morse, Feschbach, Methods of theoretical physics, ch.4.8, p.456 in proving the Parseval formula: http://www.astrosen.unam.mx/~aceves/Metodos/ebooks/morse_feshbach1.pdf
Distributions (like Dirac's $\delta$) are defined as continuous linear functionals over the set of compactly supported infinitely differentiable functions, $C_c^\infty(\mathbb R)$. Limits are defined in a weak sense.
Therefore, take $\phi \in C_c^\infty(\mathbb R)$. Then $$ I_\epsilon := \left\langle \frac{1}{\epsilon\sqrt{2\pi}} e^{-\frac{1}{2}\frac{x^2}{\epsilon^2}}, \phi(x) \right\rangle = \frac{1}{\epsilon\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}\frac{x^2}{\epsilon^2}} \phi(x) \, dx = \{ y = x/\epsilon \} \\ = \frac{1}{\epsilon\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2} \phi(\epsilon y) \, \epsilon \, dy = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2} \phi(\epsilon y) \, dy $$
Now, by the dominated convergence theorem, where we take $e^{-\frac{1}{2}y^2} \max\phi$ as the dominating function, we can take limits inside the integral: $$ I_\epsilon \to \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2} \phi(0) \, dy = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2} \, dy \, \phi(0) \\ = \phi(0) = \langle \delta, \phi \rangle. $$
This shows, $$\lim_{\epsilon\to 0} \frac{1}{\epsilon\sqrt{2\pi}} e^{-\frac{1}{2}\frac{x^2}{\epsilon^2}} = \delta(x).$$