Proof for the following nascent delta function Let $f(x)$ be a (nonzero) rapidly decreasing function in $\mathbb{R}$ (i.e. $f(x)$ and all its derivatives go to zero as $x\to\pm\infty$ faster than any negative power of $x$). The sequence of functions
$$
\eta_\epsilon(x)=\frac{\epsilon f(x)}{[\epsilon \pi  f(x)]^2+x^2}
$$
approximates the delta distribution $\delta(x)=\lim_{\epsilon\to 0}\eta_\epsilon(x)$. 
Any idea for proving of this?
It is similar to the well-known example $\frac{\epsilon}{\pi(\epsilon^2+x^2)}$, but not the same.
 A: Use the practical definition of the Dirac function below:
$$ \delta(x)=\{ \begin{matrix}
 0, & x\ne0\\ \infty, & x=0
\end{matrix}
$$
$$\int_{-\infty}^{\infty}\delta(x)dx=1$$
It is straightforward to verify the first condition and it remains to show that 
$$ \lim_{\epsilon\to 0}\int_{-\infty}^{\infty} \eta_\epsilon(x)dx=\lim_{\epsilon\to 0}I_\epsilon= 1$$
where,
$$I_\epsilon = \int_{-\infty}^{\infty} \eta_\epsilon(x)dx =\int_{-\infty}^{\infty}\frac{\epsilon f(x)dx}{[\epsilon \pi  f(x)]^2+x^2}$$
With the variable transformation,
$$ u = \frac{x}{\epsilon \pi  f(x)}$$
The integral $I_\epsilon$ becomes
$$I_\epsilon = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{du}{1+u^2}
+\epsilon \int_{-\infty}^{\infty} \frac{xf'(x)dx}{[\epsilon \pi  f(x)]^2+x^2}$$
The second integral simply vanishes in the limit $\epsilon \rightarrow 0$. Therefore, 
$$\lim_{\epsilon\to 0}I_\epsilon=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{du}{1+u^2}
=\frac{1}{\pi}\tan^{-1}(u)|_{-\infty}^{\infty}=1$$
Thus, $\lim_{\epsilon\to 0}\eta_\epsilon(x)=\delta(x)$ indeed approximates the delta function. 
