How to show that for any $r>0$ , $\lim_{n\rightarrow\infty}\int_{\mathbb{R}\backslash[-r,r]}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}=0 $ I'm trying to show that that sequence $\sqrt{\frac{n}{\pi}}e^{-nx^{2}}$ converges to dirac delta function. Proving it, I need to show that for any $r>0$ $$
 , \lim_{n\rightarrow\infty}\int_{\mathbb{R}\backslash[-r,r]}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}=0$$
So, I have
$$\lim_{n\rightarrow\infty}\int{}_{\mathbb{R}\backslash[-r,r]}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}=\lim_{n\rightarrow\infty}\left(\int_{\mathbb{R}}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}-\int_{[-r,r]}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}\right)\\=\lim_{n\rightarrow\infty}\left(1-\int_{-r}^{r}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}\right)\\=\lim_{n\rightarrow\infty}\left(1-\int_{-r\sqrt{n}}^{r\sqrt{n}}\frac{1}{\sqrt{\pi}}e^{-t^{2}}dt\right)$$
Intuitively, It is zero. But How can I show it rigorously?
 A: Hint. Since 
$$
\int_{-\infty}^{\infty}\frac{1}{\sqrt{\pi}}e^{-t^{2}}dt=1
$$ then we have
$$
\lim_{n\rightarrow\infty}\left(1-\int_{-r}^{r}\sqrt{\frac{n}{\pi}}e^{-nx^{2}}\right)=\lim_{n\rightarrow\infty}\left(1-\int_{-r\sqrt{n}}^{r\sqrt{n}}\frac{1}{\sqrt{\pi}}e^{-t^{2}}dt\right)=0.
$$
A: In most cases, you can explicitly bound the integrand by integrable functions so that you can integrate and then take the limit, especially if you want to prove the limit is zero.
$\def\nn{\mathbb{N}}$
$\def\rr{\mathbb{R}}$
$\def\less{\smallsetminus}$
Given $r > 0$ and as $n \in \nn \to \infty$:
  Thus $\int_{\rr\less[-r,r]} \sqrt{\frac{n}{π}} e^{-nx^2}\ dx \le 2 \int_r^\infty \sqrt{\frac{n}{π}} \frac{x}{r} e^{-nx^{2}}\ dx = \left[ - \sqrt{\frac{n}{π}} \frac{1}{rn} e^{-nx^2} \right]_r^\infty \to 0$.
I'll leave the last limit for you to check, but it's kind of obvious. Note that it is crucial that $r > 0$.
A: A probabilistic approach. 
$$ f_n(x) = \sqrt{\frac{n}{\pi}} e^{-nx^2} $$
is the density function of a random variable $X_n$  that is normally distributed, with mean zero and variance $\sigma_n^2 = \frac{1}{2n}$. By Chebyshev's inequality,
$$ \int_{\mathbb{R}\setminus[-r,r]}f_n(x)\,dx = \mathbb{P}[|X_n|>r\sqrt{2n}\sigma_{n}]\leq \frac{1}{2nr^2} $$
and the RHS tends to zero as $n\to +\infty$ for any $r\in\mathbb{R}^+$.
