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 $\pi$
 A: 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
$$
\begin{aligned}
\int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&=\int_{-\infty}^{\infty}\varphi(x) d\tan^{-1}
\left(\frac{x}{\varepsilon}\right) \\
&=
\left[
  \varphi(x)\tan^{-1}
\left(\frac{x}{\varepsilon}\right)
\right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\tan^{-1}
\left(\frac{x}{\varepsilon}\right)d\varphi
\end{aligned}
$$
then we have
$$
\begin{aligned}
\lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&=
-\lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\tan^{-1}
\left(\frac{x}{\varepsilon}\right)d\varphi \\
&=-\frac{\pi}{2}\int_{0}^{\infty}d\varphi+\frac{\pi}{2}\int_{-\infty}^{0}d\varphi \\
&=-\frac{\pi}{2}
\Big[
  \varphi(x)
\Big]_{0}^{\infty}+
\frac{\pi}{2}
\Big[
  \varphi(x)
\Big]_{-\infty}^{0} \\
&=\pi\varphi(0)
\end{aligned}
$$
then
$$\lim_{\varepsilon\to0}\frac{\varepsilon}{x^{2}+\varepsilon^{2}}=\pi\delta(x)$$
in the distributional sense.
A: Note that in the classical sense
$$\lim_{\epsilon \to 0}\frac{\epsilon}{x^2+\epsilon^2}=\begin{cases}0&,x\ne 0\\\\\text{undefined}&,x=0\end{cases}$$
In the sense of generalized functions, the expression 
$$\lim_{\epsilon \to 0}\frac{\epsilon}{x^2+\epsilon^2}\sim \pi \delta(x)$$
means that for all suitable test functions $\phi$ we have
$$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\phi( x)\,dx=\pi \phi(0) \tag 1$$
To show that $(1)$ is correct we let $\phi$ be a suitable test function.  Then, we can write
$$\begin{align}
\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\phi(x)\,dx&=\phi(0)\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx+\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\left(\phi(x)-\phi(0)\right)\,dx\\\\
&=\pi\,\phi(0)+\int_{-\infty}^\infty \left(\frac{1}{x^2+1}\right)\,\left(\phi(\epsilon x)-\phi(0)\right)\,dx
\end{align}$$
The Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{\epsilon \to 0}\int_{-\infty}^\infty \left(\frac{1}{x^2+1}\right)\,\left(\phi(\epsilon x)-\phi(0)\right)\,dx&=\int_{-\infty}^\infty \left(\frac{1}{x^2+1}\right)\,\lim_{\epsilon \to 0} \left(\phi(\epsilon x)-\phi(0)\right)\,dx\\\\
&=0
\end{align}$$ 
Therefore, we find that for all test functions $\phi(x)$ 
$$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,\phi( x)\,dx=\pi \phi(0)$$
as was to be shown!
A: 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 $a>0, b>0$, then
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - \pi/2 =0$$
Similary, for $a<0, b<0$, we have
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=(-\pi/2) - (-\pi/2) =0$$
For $a<0, b>0$, we have
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - (-\pi/2) =\pi$$
Hence
$$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\pi\delta(x)$$
