Understanding a proof why $\lim_{\epsilon\downarrow 0}\frac{1}{\epsilon\sqrt{2\pi}}e^{\frac{-x^2}{2\epsilon^2}} = \delta(x)$

Question

I could be making my life a bit too hard by raising this question, but the proof given by md2perpe on why $\lim_{\epsilon\downarrow 0}\frac{1}{\epsilon\sqrt{2\pi}}e^{\frac{-x^2}{2\epsilon^2}} = \delta(x)$, link: https://math.stackexchange.com/a/2834000/820472 is perfectly clear to me, except for the fact that in the end we obtain that $\lim_{\epsilon\downarrow 0}I_\epsilon = \psi(0)$. This is precisely $\left<\delta(0),\psi\right>$, however, we were tasked with showing that the said limit shows that the Dirac's delta given at any point $x$ is the weak limit of a certain Gaussian. Is it just implicitly assumed that one can reiterate the given proof with a modified Gaussian $\frac{1}{\epsilon\sqrt{2\pi}}e^{\frac{-(x - t)^2}{2\epsilon^2}}, t \in \mathbb{R}$, which then shows the equality at every $t \in \mathbb{R}$?