I hope to show that: $$\delta^3(\vec{x}-\vec{a})=\lim_{\alpha\to0} \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}$$
I want to show by: $$\int^{+\infty}_{-\infty} f(\vec{x}) \lim_{\alpha\to0} \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}\, d\vec{x}=f(\vec{a})$$ $$\lim_{\alpha\to0}\int^{+\infty}_{-\infty} f(\vec{x}) \frac{\alpha/\pi^2}{(\alpha^2+|\vec{x}-\vec{a}|^2)^2}\, d\vec{x}=f(\vec{a})$$
But I do not know how to proceed. Thank you very much.
In THIS ANSWER, I provided a rigorous development to show that the Dirac Delta can be regularized as
$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$
where
$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} $$
Here, we will use a formal approach that can easily be made rigorous as in the aforementioned post.
The interpretation of the regularization of the Dirac Delta given by
$$\delta(\vec r-\vec a)\sim \lim_{\alpha \to 0^+}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}$$
is that for any smooth test function $f(\vec r)$, we have
$$\lim_{\alpha \to 0^+}\int_{\mathbb{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV=f(\vec a)$$
First, we let $\delta>0$ and $B(\vec a,\delta)$ be a sphere of radius $\delta$ centered at $\vec a$.
Then, we split the integral into the sum of integrals over $\mathbb{R^3}\setminus B(\vec a,\delta)$ and $B(\vec a,\delta)$ and write
$$\begin{align} \int_{\mathbb{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV&=\int_{\mathbb{R}^3\setminus B(\vec a,\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV\\\\ &+\int_{B(\vec a,\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV \end{align}$$
Note that as $\alpha \to 0^+$, the first integral vanishes.
Heuristically, for "small" $\delta$, the second integral can be approximated by exploiting the continuity of $f$. Then, we find that
$$\begin{align} \lim_{\alpha \to 0^+}\int_{\mathbb{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV&=\lim_{\alpha \to 0^+}\int_{B(\vec a,\delta)}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}f(\vec r)\,dV\\\\ &=f(\vec a)\lim_{\alpha \to 0^+}\frac{\alpha}{\pi^2}\int_0^{2\pi}\int_0^\pi \int_0^{\delta}\frac{1}{(\alpha^2+r^2)^2}r^2\sin(\theta)\,dr\,d\theta\,d\phi\\\\ &=f(\vec a)\lim_{\alpha \to 0^+}\left(\frac{\alpha}{\pi^2}\,4\pi\,\left(\frac{\arctan(\delta/\alpha)}{2\alpha}-\frac{\delta}{2(\delta^2+\alpha^2)}\right)\right)\\\\ &=f(\vec a) \end{align}$$
as was to be shown!
Therefore, we have for any smooth test function $f(\vec r)$
$$\bbox[5px,border:2px solid #C0A000]{\lim_{\alpha \to 0^+}\int_{\mathbb{R}^3}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}\,f(\vec r)\,dV=f(\vec a)}$$
and it is in this sense that
$$\bbox[5px,border:2px solid #C0A000]{\delta(\vec r-\vec a)\sim \lim_{\alpha \to 0^+}\frac{\alpha /\pi^2}{(\alpha^2+|\vec r-\vec a|^2)^2}}$$
This can be done using the Fourier transform. Take the convention $$ \widehat{f}(x) = \mathcal{F}(f)(x) = \int_{\Bbb R^3} e^{-2i\pi x\cdot y} \,f(y)\,\mathrm d y \, . $$ Then (see e.g. here) $$ \frac{\alpha/\pi^2}{\bigl(\alpha^2 + |x|^2\bigr)^2} = \mathcal F(e^{-2\pi\alpha\,|y|}) $$ and so by the properties of the Fourier transform, the following limit in the sense of distributions hold $$ \frac{\alpha/\pi^2}{\bigl(\alpha^2 + |x-a|^2\bigr)^2} = \mathcal F(e^{2\pi(i\,a\cdot y - \alpha\,|y|)}) \underset{\alpha\to 0}\to \mathcal F(e^{2i\pi\,a\cdot y}) = \delta(x-a) \, . $$