Delta function from a limit?

Question

I'm having trouble understanding how the delta function comes from a specific limit I encountered in this article: at the end of page 5 and beginning of page 6 they give the following functions: $$Q(x,y,\tau)=exp\Biggl(-\frac{(y-\mu)^2}{2\Sigma^2}\Biggr)$$ with $\mu=xe^{-b\tau}+\frac{a}{b}(1-e^{-b\tau})$ e $\Sigma^2=\frac{\sigma^2}{2b}(1-e^{-2b\tau})$ and $$\phi(y,\tau)=\frac{1}{\sqrt{1-e^{-2b\tau}}}$$ Then they claim that in order to fit to the intial condition for $f(x,y,\tau)=\phi(y,\tau)Q(x,y,\tau)$ that is $f(x,y,0)=\delta(x-y)$, one can do the limit of $f$ for $\tau$ that tends to $0$ and find the missing term that should be $(2\pi\sigma^2/(2b))^{-1/2}$ so that $$f(x,y,\tau)=\frac{1}{\sqrt{2\pi}\Sigma}exp\Biggl(-\frac{(y-\mu)^2}{\Sigma^2}\Biggr)$$ I really don't understand how to get that term from the limit, can anyone explain this step?