As part of a proof of the Fourier Inversion Theorem in $\mathbb{R}^N$ I want to show that $$\lim_{\epsilon\to \infty}\frac{1}{(2\pi)^{N/2}\epsilon^N}\int_{\mathbb{R}^N} e^{-\frac{|x/\epsilon|^2}{2}} f(x) dL^N(x) = f(0).$$
My convention for the Fourier inversion is $\hat{f}(\xi) = \frac{1}{(2\pi)^{N/2}}\int_{\mathbb{R}^N} f(x)e^{-i\xi\cdot x} dL^N(x) $.
$f$ is assumed continuous at $x = 0$
I honestly don't know where to start. I feel like I want to pull the $\lim$ into the integral sign (using some convergence theorem perhaps?) and then maybe look at the expression $$\lim_{\epsilon \to \infty} \frac{e^{-i\xi\cdot x}}{\epsilon^N}.$$
Start with changing the variable in the integral $x=\epsilon y$. $f(x)$ will be replaced by $f(\epsilon y)$, which will pointwise tend to $f(0)$ by continuity of $f$, the Jacobian will take care of the $\epsilon^{-N}$, a suitable convergence theorem will be invoked (e.g. Lebesgue's dominated convergence theroem) and everything will be OK.