Question: Find the limit of the function $f(x) = \lim_{\epsilon \rightarrow 0^+}\frac{e^\frac{-x^2}{\epsilon}}{\sqrt \epsilon}$ in the distribution sense:
My Attempt : $\left< f_\epsilon,\phi \right> = \int_{-\infty}^{\infty}\frac{e^\frac{-x^2}{\epsilon}}{\sqrt \epsilon}\phi(x) dx= -2\int_{-\infty}^{\infty}\frac{xe^\frac{-x^2}{\epsilon}}{\epsilon^\frac{3}{2}}\phi(x) dx$
I was planning to use the change of variable technique. But stucked. Could you please solve this problem for me ?
On
$$\begin{align} \langle f_\epsilon, \phi \rangle &= \epsilon^{-1/2} \int_{-\infty}^{\infty} e^{-x^2/\epsilon} \, \phi(x) \, dx \\ &= \{ \xi = \epsilon^{-1/2} x \} \\ &= \int_{-\infty}^{\infty} e^{-\xi^2} \, \phi(\epsilon^{1/2} \xi) \, d\xi \\ &\to \int_{-\infty}^{\infty} e^{-\xi^2} \, \phi(0) \, d\xi \\ &= \left( \int_{-\infty}^{\infty} e^{-\xi^2} \, d\xi \right) \phi(0) \\ &= \{ \text{Gaussian integral} \} \\ &= \sqrt{\pi} \, \phi(0) \\ &= \langle \sqrt{\pi} \, \delta, \phi \rangle \end{align}$$
See Wikipedia about the Gaussian integral.
$$\lim_{\epsilon \rightarrow 0^+}\frac{e^\frac{-x^2}{\epsilon}}{\sqrt \epsilon} = \sqrt{\pi} \lim_{\epsilon \rightarrow 0^+}\frac{e^\frac{-x^2}{2 \epsilon}}{\sqrt{ 2\pi \epsilon}} $$
Limit of standard normal distribution is Dirac function at zero. Hence the limit above is $\sqrt{\pi} \delta(x).$