Show that $\lim_{a\to 0^+} \int \frac{e^{-|x|/a}}{2a}f(x)dx=f(0)$ I'm trying to show $\displaystyle \lim_{a\to 0^+} \int \frac{e^{-|x|/a}}{2a}f(x)dx=f(0)$ where $f$ is continuous with compact support. 
I've already shown that for any $a>0, \displaystyle\int\frac{e^{-|x|/a}}{2a}dx=1$ and that for any fixed $\delta>0$, $\displaystyle\lim_{a\to 0^+}\int_{|x|>\delta}\frac{e^{-|x|/a}}{2a}dx=0$.
Since the limit is supposed to be $f(0)$, I assume I'll need to use some sort of integration by parts/Fundamental Theorem of Calculus. None of the convergence theorems seem helpful since, at $x=0$, $\displaystyle\lim_{a\to 0^+}\frac{e^{0}}{2a}=\infty$. Is there a theorem or technique that I should consider for this? Thank you 

Do you know about mollifiers? If you do, the last thing to see is that
$$\frac{e^{-|x|/a}}{2a} \ge 0 \qquad \forall a > 0$$
And then notice that
$$\int \underbrace{\frac{e^{-|x|/a}}{2a}}_{=: \phi_a} f(x) \mathrm dx = (\phi_a \ast f)(0)$$
Then use that $(\phi_a)_{a>0}$ is a mollifier famliy and use
$$\lim_{a\searrow 0} (\phi_a \ast g)(x) = g(x)$$
in all points of continuity (this works even for non-compactly supported $g$).

Using this result, the conditions on $f$ can be significantly weakened to $f\in L^1(\mathbb R)$ and $f$ is continuous at $0$.
