prove $\frac{n e^{-n^2 x^2}}{\sqrt{\pi }}$ converges to $\delta(x)$ How can show as $n$ goes toward infinity the sequence converges to $\delta(x)$ my problem is I don't know how to show this.  
 A: Convergence in the sense of distributions is defined as follows: a sequence of distributions $\tau_n\in \mathscr D'(\mathbb R)$ is said to converge to a distribution $\tau\in\mathscr D'(\mathbb R)$ if, for every test function $\varphi\in C^{\infty}_0(\mathbb R)$, there holds
$$
\lim_{n\to\infty}\langle \tau_n, \varphi\rangle =\langle \tau, \varphi\rangle,
$$
in the usual topology of $\mathbb R$. The $\langle\cdot,\cdot\rangle$ denotes the action of the distribution on the test function:
$$
\tau:C^{\infty}_{0}(\mathbb R) \longrightarrow \mathbb R\\
\varphi \longmapsto \langle \tau, \varphi \rangle,
$$
which is linear and continuous in the appropriate sense. Of course, distributions which are in fact locally summable functions $\tau_f\simeq f$ act on test functions $\varphi$ by
$$
\langle \tau_f, \varphi \rangle = \int_{\mathbb R}f(x) \varphi(x) dx.
$$
Consider a test function $\varphi\in C^\infty_0(\mathbb R)$ (i.e. infinitely regular and of compact support). Then letting
$$
\rho_n(x)\equiv \frac{n}{\sqrt{\pi}}e^{-n^2x^2}
$$
we have
$$
\langle \rho_n, \varphi \rangle = 
\int_{-\infty}^{+\infty} \frac{n}{\sqrt{\pi}}e^{-n^2x^2} \varphi(x) dx\\
=\int_{-\infty}^{+\infty} \frac{1}{\sqrt{\pi}}e^{-y^2} \varphi(y/n) dy
$$
where we have used the change of variables $y=nx$ in the last step. Since the integrand is dominated by the (summable) function
$$
e^{-y^2} \sup_{x\in\text{supp}\varphi}|\varphi(x)|
$$
we can apply Lebesgue's theorem and pull the limit $n\to\infty$ inside the integral getting
$$
\lim_{n\to\infty}\langle \rho_n, \varphi \rangle =
\int_{-\infty}^{+\infty} \frac{1}{\sqrt{\pi}}e^{-y^2} \varphi(0) dy = \varphi(0)
$$
since $\int_{\mathbb R}e^{-y^2}dy = \sqrt{\pi}$.
So we are ready to check the definition: since for every such $\varphi$
$$
\lim_{n\to\infty}\langle \rho_n, \varphi\rangle =\varphi(0)= \langle\delta, \varphi\rangle,
$$
then
$$
\lim_{n\to\infty}\rho_n = \delta
$$
in the sense of distributions over $C^{\infty}_0(\mathbb R)$.
