prove that $\lim_{n\to\infty}\int_{-\infty}^\infty \frac n {\sqrt {\pi}}e^{-(nx)^2} f(x) dx = f(0)$ prove that for any integrable function $f$ defined on $(-\infty,\infty)$ (or maybe square integrable function.. my professor didn't gave exact imformation about $f$.)
$$\lim_{n\to\infty}\int_{-\infty}^\infty \frac n {\sqrt {\pi}}e^{-(nx)^2} f(x) dx = f(0)$$
I think I have to get some $\delta>0$ and divide integral into three terms.  
$$\int_{-\infty}^{-\delta} f*g_n +\int_{-\delta}^{\delta}f*g_n + \int_\delta ^\infty f*g_n$$
($g_n(x)=\frac n {\sqrt {\pi}}e^{-(nx)^2}$)
since $f*g_n$ converges uniformly to $0$ if $x\neq0$, first and third term vanish.
so what I have to do is to take $N$ and $\delta$ (from the given value $\epsilon>0$) so that if $n>N$  $$\left|\int_{-\delta}^{\delta}\frac n {\sqrt {\pi}}e^{-(nx)^2} f(x)dx - f(0) \right|<\epsilon$$
I can't go any further.
 A: Hints: 1. Since we could redefine $f$ at the one point $0$ and not change any of the integrals, we need further conditions on $f$ for this to be true. For example, continuity of $f$ at $0.$ 2. This is not a DCT problem. 3.This problem is but one example of the following situation: We have a sequence integrable functions $g_n$ on $\mathbb R$ such that i)$\int_{\mathbb R}g_n = 1$ for all $n;$ ii)for all $r>0,$  $g_n(x)\to 0$ uniformly on $\{|x|>r\}.$
If we have that, then given any integrable $f$ on $\mathbb R$ that is continuous at $0,$
$$\lim_{n\to \infty} \int_{-\infty}^\infty g_n\,f = f(0).$$
Generalizing like this actually makes the proof easier I think.
A: Not an answer, but too long for a comment.
Using a substitution, the left-hand side can be rewritten as $\lim_{n\to\infty}\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}f\left(\frac{u}{n}\right) du$. The first hard part is proving the integration commutes with the limit-taking, so that we can rewrite again to give $\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}\lim_{n\to\infty}f\left(\frac{u}{n}\right) du$. For continuous $f$, this is $$\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}f\left(\lim_{n\to\infty}\frac{u}{n}\right) du=\int_{-\infty}^\infty\frac{1}{\sqrt{\pi}}e^{-u^2}f\left(0\right) du=f\left(0\right).$$As integrable functions are not in general continuous, what can be said about integrable $f$ is the second hard part.
A: This is my try. The continuity at $0$ and the boundedness of $f$ should be enough.
Rewrite the LHS as: 
$$\int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi\frac{1}{2n^2}}} e^{-\frac{x^2}{2\frac{1}{2n^2}}} f(x)dx$$
This is the expectation of function $f(X_n)$, where $X_n \sim N(0, \frac{1}{2n^2})$. We need to show that: $E(f(X_n)) \to f(0)$.
We can show that $X_n \overset{a.s.}{\longrightarrow} 0$. As $f$ is continuous at $0$, we have: $f(X_n) \overset{a.s.}{\longrightarrow} f(0)$. $f(X_n)$ is bounded, so we can use DCT to conclude.
