Prove that $$g_\epsilon (x)=\lim_{\epsilon \to 0} \frac1 \epsilon \frac1 \pi e^{-x^2/\epsilon^2}$$ is a Dirac-$\delta$ function.
This is a homework question I'm stuck with. I'm probably missing a very simple point, and can't seem to figure it out. Any help to prompt me in the right direction would be much appreciated.
What I've done so far is the following.
I need to show that, $$\int_{-\infty}^\infty f(t)g_\epsilon (x)dt=f(0)$$ So let $t=\epsilon x$, $$\lim_{\epsilon \to 0} \int_{-\infty}^\infty f(t) \frac1 \epsilon e^{-x^2/\epsilon^2}dt=$$ $$\stackrel{t=\epsilon x}{=}\lim_{\epsilon \to 0} \int_{-\infty}^\infty f(\epsilon x) \frac1 \epsilon \frac1 \pi e^{-x^2/\epsilon^2}\epsilon dx$$ $$=f(0) \frac1 \pi \lim_{\epsilon \to 0} \int_{-\infty}^\infty e^{-x^2/\epsilon^2} dx$$ And since $\lim_{\epsilon \to 0} \int_{-\infty}^\infty e^{-x^2/\epsilon^2}=0$, the result is $0$, not $f(0)$.
Another approach was to just set $t=x$ so that $\lim_{\epsilon \to 0} \int_{-\infty}^\infty \frac1 \epsilon e^{-x^2/\epsilon^2}=\sqrt\pi$ would cancel out with $1/\pi$, but then I'm left with $f(x)$ instead of $f(0)$.
As Jordan pointed out in a comment, if we demand $\int_{-\infty}^{\infty} f(t) g_{\epsilon} (t) dt = f(0)$ and let $t=\epsilon x$ then
$$\int_{-\infty}^{\infty} f(t) g_{\epsilon} (t) dt = \lim_{\epsilon\to 0} \int_{-\infty}^{\infty} f(t) \frac{1}{\epsilon\sqrt{\pi}}e^{-t^2/\epsilon^2}dt =\\ \lim_{\epsilon\to 0} \int_{-\infty}^{\infty} f(\epsilon x) \frac{1}{\epsilon\sqrt{\pi}}e^{-x^2} \epsilon dx = \int_{-\infty}^{\infty} f(0) \frac{1}{\sqrt{\pi}}e^{-x^2} dx =\\ f(0) \frac{1}{\sqrt{\pi}} \cdot \sqrt{\pi} = f(0) $$
P.S. I think you need $\sqrt{\pi}$ when defining $g_\epsilon$.