Dirac identity of Gaussian Heat Kernel I am having a real hard time trying to figure this problem out:
$$u(\textbf{x},t)=\int_{\mathbb{R}^n}K(\textbf{x},\textbf{y},t)g(\textbf{y})d\textbf{y}$$
Where:
$$ K(\textbf{x},\textbf{y},t):= \frac{e^{-\frac{|\textbf{x}-\textbf{y}|}{4t}^2}}{(4\pi t)^{n/2}}
$$
show that
$$
\lim_{x\to 0}u(\textbf{x},t)=g(\textbf{x})
$$
so far I have shown that 
$$
\int_{\mathbb{R}^n}K(\textbf{x},\textbf{y},t)d\textbf{y}=1
$$
But I can't seem to figure out how to deal with the limit since you have an infinity over infinity that can't seem to be solved using L'Hospital's rule. I get the general notion, namely that as t $\to 0$, the Gaussian Heat Kernel acts like the Dirac Delta function, but I can't seem to prove it. 
Thanks
 A: 
HINT:

First, we have $\lim_{t\to 0^+}\frac{e^{-|\vec x-\vec y|^2/(4t)}}{(4\pi t)^{n/2}}=0$ for $|\vec x-\vec y|\ne 0$.  So, we need to isolate the singularity point, $\vec x$, from the integration.  
Proceeding, we write the integral as
$$\begin{align}
u(\vec x,t)&=\int_{\mathbb{R}^n}K(\vec x,\vec y,t)g(\vec y)\,d\vec y\\\\
&=\int_{\mathbb{R}^n\setminus B(\vec x,\delta)}K(\vec x,\vec y,t)g(\vec y)\,d\vec y+\int_{B(\vec x,\delta)}K(\vec x,\vec y,t)(g(\vec y)-g(\vec x))\,d\vec y\\\\
&+g(\vec x)\int_{B(\vec x,\delta)}K(\vec x,\vec y,t)\,d\vec y\\\\
\end{align}$$
where $B(\vec x,\delta)$ is a ball of radius $\delta$ centered at $\vec x$.

Moreover, we have
$$\begin{align}
\lim_{t\to 0^+}\int_{B(\vec x,\delta)}K(\vec x,\vec y,t)\,d\vec y&=\lim_{t\to 0^+}\left(\frac{1}{(4\pi t)^{n/2}}\int_{S^{n-1}}\int_{0}^\delta e^{-r^2/4t}\,r^{n-1}\,dr\,\,dS^{n-1}\right)\\\\
&=\frac{2}{\Gamma(n/2)}\,\lim_{t\to 0^+}\int_0^{\delta/(2\sqrt t)}e^{-r^2}r^{n-1}\,dr\\\\
&=1
\end{align}$$


Can you proceed to fill in the rest?


Alternatively, translate coordinates by letting  $\vec y-\vec x=\vec y_{\text{new}}$ so that $g(\vec y)=g(\vec x+\vec y_{\text{new}})$, scale the radial variable $r=|\vec y_{\text{new}}| \to r/(2\sqrt t)$, and simply apply the Dominated Convergence Theorem to bring the limit inside the integral.  The result is $g(\vec x)\frac{2}{\Gamma(n/2)}\int_{\mathbb{R}^n}e^{-r^2}\,r^{n-1}\,dr=g(\vec x)$ as expected.
