approximating to delta sequence I want to prove the following: 
Let $h_{n}(t)=\sqrt {n}e^{-nt^{2}}$. The area under the graph of $h_{n}(t)$ is $\sqrt {\pi}$ and for any $\epsilon>0$ $h_{n}(t) \rightarrow0$ uniformly outside $]-\epsilon, \epsilon [$. 
Show that if $g(t)$ is continuous and $0<N< \infty$ then $\int_{-N}^{N} g(t)h_{n}(t)dt \rightarrow g(0)\sqrt{\pi} $ 
I try to use the fact that we can approach to the function g by sequence of polynomials but I couldnt get the result. Any help would be great. Thanks
 A: Let
$$I_n = \int_{-N}^{N}g(t) \sqrt{n}e^{-nt^2}\,dt.$$
Note that
$$g(0)\int_{-\infty}^{\infty} \sqrt{n}e^{-nt^2}\,dt= \sqrt{\pi}g(0).$$
Hence, using the change of variables $u = \sqrt{n}t$,
$$|I_n - \sqrt{\pi}g(0)| \\ \leqslant \int_{-N}^{N}|g(t)-g(0)| \sqrt{n}e^{-nt^2}\,dt + |g(0)|\int_{N}^{\infty} \sqrt{n}e^{-nt^2}\,dt + |g(0)|\int_{-\infty}^{-N} \sqrt{n}e^{-nt^2}\,dt \\ \leqslant \int_{-N}^{N}|g(t)-g(0)| \sqrt{n}e^{-nt^2}\,dt + |g(0)|\int_{N\sqrt{n}}^{\infty} e^{-u^2}\,du + |g(0)|\int_{-\infty}^{-N\sqrt{n}}e^{-u^2}\,du .$$
As $\displaystyle \int_{-\infty}^{\infty} e^{-u^2} \, du < \infty$, the second and third terms on the RHS are each less than $\epsilon/3$ if $n$ is sufficiently large, and
$$|I_n - \sqrt{\pi}g(0)|  \leqslant  \int_{-N}^{N}|g(t)-g(0)| \sqrt{n}e^{-nt^2}\,dt +\frac{2\epsilon}{3}.$$
Since $g$ is continuous on $[-N,N]$ it is bounded and $|g(t)| \leqslant M$. Also there exists $\delta > 0$ such that $|g(t) - g(0)| < \epsilon/9\sqrt{\pi}$ when $|t| < \delta.$
Hence,
$$\int_{-N}^{N}|g(t)-g(0)| \sqrt{n}e^{-nt^2}\,dt  \\ < \frac{\epsilon}{9\sqrt{\pi}}\int_{-\delta}^{\delta} \sqrt{n}e^{-nt^2}\,dt + 2M \int_{\delta}^{N} \sqrt{n}e^{-nt^2}\,dt + 2M \int_{-N}^{-\delta} \sqrt{n}e^{-nt^2}\,dt \\ < \frac{\epsilon}{9} + 2M \int_{\delta\sqrt{n}}^{N\sqrt{n}} e^{-u^2}\,du + 2M \int_{-N\sqrt{n}}^{-\delta\sqrt{n}} e^{-u^2}\,du.$$
If $n$ is sufficiently large then the two integrals on the RHS are each less than $\epsilon/ 18M$, and
$$\int_{-N}^{N}|g(t)-g(0)| \sqrt{n}e^{-nt^2}\,dt  < \frac{\epsilon}{3}.$$
Therefore, if $n$ is sufficiently large
$$|I_n - \sqrt{\pi}g(0)| < \epsilon.$$
