I have been struggling with the following problem and any help / good place to start would be appreciated.
Show,
$$\lim_{\epsilon \to 0} f_\epsilon (x-x_0) = \delta(x-x_0)$$
when
$$f_\epsilon = {(2\epsilon)^{-1}} for |x|\leqslant \epsilon $$
and
$$f_\epsilon = 0 \ \ \text{otherwise.} $$
A physicist approach is to notice that the function $f_{\epsilon}$ goes to infinity when $\epsilon\to 0$ while the integral over $\Bbb{R}$ is equal to $1$ and a physicist concludes that the limit is $\delta_0$.
Now I am not a physicist so let's look at it from the distribution theory angle. $f_\epsilon$ is locally intégrable so it defines a distribution that is a continuous linear operator on the test functions $\varphi\in\mathcal{D}(\Bbb{R})$ ($\mathcal{C}^\infty$ with compact support)
$$\langle f_\epsilon,\varphi\rangle=\int_{\Bbb{R}}f_\epsilon(x)\cdot \varphi(x)dx={1\over 2\epsilon}\int_{-\epsilon}^{\epsilon}\varphi(x)dx $$
for $\epsilon$ small enough.
Now we can prove that
$$\lim_{\epsilon\to 0}{1\over 2\epsilon}\int_{-\epsilon}^{\epsilon}\varphi(x)dx=\varphi(0)$$
And in distribution theory delta is the linear form $\delta_0:\varphi\to\varphi(0)$