If I have a real-value function (for $x_0 > 0$) $$ f(x; x_0) = \frac{1}{x_0}\exp\Bigg(-\frac{|x|}{x_0}\Bigg) $$ If I plot this for different values of $x_0$ it becomes more and more narrow with $x_0 \rightarrow 0$.
$f(x;x_0)$ for different values of $x_0$" />
Can we prove that this function can be used as a limit representation of Dirac-delta function in the limit $x_0 \rightarrow 0$?
i.e. Is the following true, and if it is how do we prove it?
$$ \lim_{x_0\rightarrow0}f(x;x_0)=\delta(x) $$
Let $\phi\in C_C^\infty$. Then, the Dominated Convergence Theorem guarantees that
$$\begin{align} \lim_{x\to0^+}\int_{-\infty}^\infty \left(\frac{e^{-|x|/x_0}}{2x_0}\right)\phi(x)\,dx&=\lim_{x\to0^+}\frac12\int_{-\infty}^\infty e^{-|x|}\phi(x_0x)\,dx\\\\ &=\phi(0) \end{align}$$
Therefore, in distribution we have
$$\lim_{x_0\to 0^+}\left(\frac{e^{-|x|/x_0}}{2x_0}\right)=\delta(x)$$
And we are done!