$\{e^{-x^2/n^2}\}$ converges to the generalized function 1
I could not prove this result from "M. J. Lighthill-Introduction to Fourier analysis and generalized functions (1964)" (p.17)
It is easy to prove that if $f$ is fairly good ($\exists n\in \mathbb{N} \forall m\in\mathbb{N} $ s.t. $ \lim_{|x|\rightarrow \infty} \dfrac{f^{(m)}(x)}{|x|^n}<\infty$) and $\{a_n\}\rightarrow\alpha$, then $\{fa_n\}\rightarrow f\alpha$. Putting $a_n=e^{-x^2/n^2}$, I will be able to prove the following result from "Dennery&Krzywicki-Mathematics for Physicists (1996)" p. 228:
Given a fairly good function $f$, and given a good function $g$,
$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty} e^{-x^2/n^2} f g \mathrm{d}x=\int_{-\infty}^{\infty}fg \mathrm{d}x$
Since the writer proved this in long, I assume that proving $\{e^{-x^2/n^2}\}\rightarrow 1$ is also hard.
$|e^{-x^2/n^2}f(x)g(x)| \le |f(x)g(x)| \in L^1$ for each $x,n$. Since $e^{-x^2/n^2} \to 1$ as $n \to \infty$ for each $x$, dominated convergence theorem does the job.