How to show $\lim_{\epsilon \rightarrow 0} \frac{\epsilon}{x^2+\epsilon ^2} = \pi \delta (x)$? Show that : 

$\lim_{\epsilon \rightarrow  0} \frac{\epsilon}{x^2+\epsilon ^2} = \pi  \delta (x)$

Where $\delta (x)$ is the dirac-delta function.

I can't show that the integral of this over all $x$ is $\pi$
Since you tagged distribution theory, I think the following should work.

For any test function $\varphi(x)\in\mathcal{D}(\mathbb{R})$, where $\mathcal{D}(\mathbb{R})$ consists of $C^{\infty}$-smooth functions with compact support (which indicates that $\lim_{x\to\pm\infty}\varphi(x)=0$). We have the following
$$
\begin{aligned}
\int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&amp;=\int_{-\infty}^{\infty}\varphi(x) d\tan^{-1}
\left(\frac{x}{\varepsilon}\right) \\
&amp;=
\left[
  \varphi(x)\tan^{-1}
\left(\frac{x}{\varepsilon}\right)
\right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\tan^{-1}
\left(\frac{x}{\varepsilon}\right)d\varphi
\end{aligned}
$$
then we have
$$
\begin{aligned}
\lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&amp;=
-\lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\tan^{-1}
\left(\frac{x}{\varepsilon}\right)d\varphi \\
&amp;=-\frac{\pi}{2}\int_{0}^{\infty}d\varphi+\frac{\pi}{2}\int_{-\infty}^{0}d\varphi \\
&amp;=-\frac{\pi}{2}
\Big[
  \varphi(x)
\Big]_{0}^{\infty}+
\frac{\pi}{2}
\Big[
  \varphi(x)
\Big]_{-\infty}^{0} \\
&amp;=\pi\varphi(0)
\end{aligned}
$$
then
$$\lim_{\varepsilon\to0}\frac{\varepsilon}{x^{2}+\varepsilon^{2}}=\pi\delta(x)$$
in the distributional sense.
For $x\ne 0$,
$$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=0$$
For $x=0$,
$$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\infty$$
Besides,
$$\int_a^b \frac{\epsilon}{x^2+\epsilon^2}dx=\int \frac{1/\epsilon}{1+(x^2/\epsilon^2)}dx$$
$$=\int_a^b \frac{1}{1+(x^2/\epsilon^2)}d(x/\epsilon)$$
$$=\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}$$
Now if $a&gt;0, b&gt;0$, then
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - \pi/2 =0$$
Similary, for $a&lt;0, b&lt;0$, we have
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=(-\pi/2) - (-\pi/2) =0$$
For $a&lt;0, b&gt;0$, we have
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - (-\pi/2) =\pi$$
Hence
$$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\pi\delta(x)$$