<s>[INST] Convergence of $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ as $\varepsilon \to 0$ Is it true that $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ converges (in some sense) to $\delta_0$, Dirac delta distribution at point $0$,  as $\varepsilon \to 0$ ? Thanks.
 [/INST] So the $i$ is intentional. Then let's look how that function acts as a distribution, hence let $\varphi$ an arbitrary test function:
$$\begin{align}
\frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\infty \exp \left(i\frac{x^2}{2\varepsilon}\right)\varphi(x)\,dx &= \frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\infty \left(\cos \frac{x^2}{2\varepsilon} + i\sin \frac{x^2}{2\varepsilon}\right)\varphi(x)\,dx\\
&= \frac{1}{\sqrt{\pi i}}\int_{-\infty}^\infty \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\
&= \frac{1}{\sqrt{\pi i}}\int_{-\infty}^0 \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\
&\qquad + \frac{1}{\sqrt{\pi i}}\int_{0}^\infty \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\
&= \frac{1}{\sqrt{\pi i}} \int_0^\infty \left(\cos (y^2) + i\sin (y^2)\right)\left(\varphi(\sqrt{2\varepsilon}\cdot y) + \varphi(-\sqrt{2\varepsilon}\cdot y) \right)\,dy.
\end{align}$$
Now for $\varepsilon = 0$, we have the Fresnel integral
$$\frac{2\varphi(0)}{\sqrt{\pi i}}\int_0^\infty \cos (y^2) + i \sin (y^2)\,dy = \frac{2\varphi(0)(1+i)}{\sqrt{\pi i}}\sqrt{\frac{\pi}{8}} = \varphi(0),$$
and it remains to argue that taking the limit is legitimate. I'll leave that to handwaving, it can be done.
</s>