I started doing this problem, but got stuck in the middle:
I want to find the distributional limit to $$\frac {xe^{\frac{-x^2}{s^2}}}{s^3}$$ when $s\rightarrow 0$ :
Here`s what I did: usig integration by parts
$\int_R \frac {xe^{\frac{-x^2}{s^2}}}{s^3} . \phi(x) dx=-\int_R \frac{e^{-x^2/s^2}}{s} . d(x\phi(x))= -\int_R \frac{e^{-x^2/s^2}}{s} . (\phi (x)+x\phi'(x)) dx$
how can I proceed now?
Thanks
Your distribution can be written as $$\frac {xe^{\frac{-x^2}{s^2}}}{s^3} = -\frac{d}{dx} \left(\frac {1}{2s}e^{\frac{-x^2}{s^2}}\right).$$
Then you can say that by a well-known result (ask in comments if do not know how to prove it) $$\lim_{s\to 0} \frac {1}{2s}e^{\frac{-x^2}{s^2}} = \left(\frac 12 \int_\Bbb R e^{-x^2}dx\right)\delta_0.$$
Finally, another standard result in distribution theory is that if $T_n\to T$ in $D'$, then $T'_n\to T'$ in $D'$.
edit
Proof of the second equality (up to a factor $\frac 12$).
We simply write by definition for a test function $\phi$ and a change of variables $y = x/s$: $$\left\langle\frac {1}{s}e^{\frac{-x^2}{s^2}},\phi(x) \right\rangle = \int_\Bbb R \frac {1}{s}e^{\frac{-x^2}{s^2}}\phi(x)dx = \int_\Bbb R e^{ -y^2 }\phi(ys)dy.$$ By the dominated convergence theorem when $s\to 0+$ the latter integral converges to $$\int_\Bbb R e^{ -y^2 }\left(\lim_{s\to 0+}\phi(ys)\right)dy = \int_\Bbb R e^{ -y^2 }\phi(0)dy = \left\langle \left(\int_\Bbb R e^{-y^2}dy\right)\delta_0,\phi\right\rangle$$