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 $\delta&gt;0$ 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 $\epsilon&gt;0$) so that if $n&gt;N$  $$\left|\int_{-\delta}^{\delta}\frac n {\sqrt {\pi}}e^{-(nx)^2} f(x)dx - f(0) \right|&lt;\epsilon$$

I can't go any further.
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 $r&gt;0,$  $g_n(x)\to 0$ uniformly on $\{|x|&gt;r\}.$

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.
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.