How can I prove $\lim_{\epsilon \to 0} \space \text{Im}\frac{1}{x+i \epsilon}=-\pi\delta(x)$? I want to show that:

$$\lim_{\epsilon \to 0} \space \text{Im}\frac{1}{x+i \epsilon}=-\pi\delta(x)$$

This is my attempt:


*

*I assumed that $\text{Im}$ stands for the imaginary part. Therefore, $$\frac{1}{x+i \epsilon} \cdot \frac{x-i \epsilon}{x-i\epsilon}=\frac{x-i\epsilon}{x^2+\epsilon^2}=\frac{x}{x^2+\epsilon^2}-i \left (\frac{ \epsilon}{x^2+\epsilon^2} \right ) \implies\Im(z)=-\frac{1}{\frac{x^2}{\epsilon}+\epsilon}$$

*Taking the limit: $$\lim_{\epsilon \to 0} -\frac{1}{\frac{x^2}{\epsilon}+\epsilon}=0$$


But this would mean that $$\implies0=-\pi \delta(x) $$
However, this equality can't be true because: 
$$\delta(x)=\begin{cases}\infty, \space \space\space x=0 \\ 0, \space \space \space x \not= 0 \end{cases}$$
which would mean that at $x=0$ the equality $0=-\pi \delta(x)$ is not true. What am doing wrong?
 A: 
In THIS ANSWER and THIS ONE, I provide primers on the Dirac Delta.

For any smooth test function $\phi$, we have for any $\nu >0$, there exists a $\delta>0$ such that $|\phi(x)-\phi(0)|<\nu$ whenever $|x|<\delta$.  Then, we can write
$$\begin{align}
\lim_{\epsilon \to 0}\int_{-\infty}^\infty \phi(x)\text{Im}\left(\frac{1}{x+i\epsilon}\right)\,dx&=-\lim_{\epsilon \to 0}\int_{-\infty}^\infty \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\
&=-\lim_{\epsilon \to 0}\int_{|x|\ge \delta} \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx-\lim_{\epsilon \to 0}\int_{-\delta}^\delta \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\
&-\lim_{\epsilon \to 0}\int_{-\delta}^\delta \phi(x)\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\
&=-\lim_{\epsilon \to 0}\left(\int_{-\delta}^\delta (\phi(x)-\phi(0))\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx-\phi(0)\int_{-\delta}^\delta \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\right)\\\\
&=-\pi \phi(0)-\lim_{\epsilon \to 0}\int_{-\delta}^\delta (\phi(x)-\phi(0))\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\\\\
\end{align}$$
Note that we have for any $\nu>0$
$$\left|\int_{-\delta}^\delta (\phi(x)-\phi(0))\left(\frac{\epsilon}{x^2+\epsilon^2}\right)\,dx\right|\le 2\nu \arctan(\delta/\epsilon)\le \pi \nu $$
Therefore, for any smooth tests function $\phi$, we find 
$$\lim_{\epsilon \to 0}\int_{-\infty}^\infty \phi(x)\text{Im}\left(\frac{1}{x+i\epsilon}\right)\,dx=-\pi \phi(0)$$
This is equivalent to the statement that 
$$\lim_{\epsilon \to 0}\text{Im}\left(\frac{1}{x+i\epsilon}\right)\sim -\pi \delta(x)$$
in the sense of a regularization of the Dirac Delta.
