What is the limit $\lim_{s \to 0}\frac{1}{\pi s^2} e^{-r^2/s^2}$

Question

There is a typo in one of the papers I just read and instead of the known delta function limit

$\lim_{s \to 0}\frac{1}{\pi s} e^{-r^2/s^2}$

it says

$\lim_{s \to 0}\frac{1}{\pi s^2} e^{-r^2/s^2}$

Note the $1/s^2$ instead of the $1/s$. I was wondering what that limit might yield if applied to a proper test function and integrated.

Answer 1

Let $\phi$ a test function such that $\phi(0) = 1$ then $$\frac{1}{\pi s^2} \int_{\mathbb R}e^{-\frac{r^2}{s^2}} \phi(r)dr = \frac{1}{s}\frac{1}{\pi s} \int_{\mathbb R}e^{-\frac{r^2}{s^2}} \phi(r)dr $$ and if $s\rightarrow 0^+$ $$\frac{1}{\pi s} \int_{\mathbb R}e^{-\frac{r^2}{s^2}} \phi(r)dr \rightarrow \phi(0) = 1$$ $$\frac{1}{s} \rightarrow \infty$$ so $$\frac{1}{\pi s^2} \int_{\mathbb R}e^{-\frac{r^2}{s^2}} \phi(r)dr \rightarrow \infty $$ so it doesn't converge.

Answer 2

Let $u_s=\frac{1}{\pi s} e^{-r^2/s^2}$. Observe that if $\frac{1}{s} u_s\rightarrow u$ as $s\rightarrow 0$, then $u_s\rightarrow 0$ as $s\rightarrow 0$. It follows, since $u_s\rightarrow \delta$ and $\delta\neq 0$, that $\frac{1}{s}u_s$ does not converge.