We know that $$\lim_{\varepsilon \to 0} \frac{1}{\pi} \frac{\varepsilon}{\varepsilon^2 + x^2} = \delta(x).$$
This is because the function has a constant unit integral over $\mathbb{R}$ and is increasingly localized as we approach the limit.
If we consider the functions $\eta_{\varepsilon} : \mathbb{R}^2 \mapsto \mathbb{R}$ $$\eta_{\varepsilon}(\mathbf{r}) = \frac{\varepsilon}{\varepsilon^2 + \mathbf{r}\cdot \mathbf{r}},$$ we have a diverging integral over $\mathbb{R}^2$ for finite $\varepsilon$, so the limit is indeterminate.
However, in polar coordinates, we could write the $\eta_{\varepsilon}$ function as depending only on the radial variable. Can we then formally take the following limit? \begin{align} \lim_{\varepsilon \to 0} \int_{0}^{2\pi} \mathrm{d}\phi \int_{0}^{\infty} \frac{\varepsilon r f(r)}{\varepsilon^2+r^2} \mathrm{d}r = 2\pi rf(r) \Big|_{r=0}, \end{align} which also solves the indeterminacy in the limit for the integral of $\eta_{\varepsilon}$ concluding that it's zero.
Edit: Thanks to @Klaus I think that the question can be formulated differently. For the functions $\zeta_{\varepsilon} : \mathbb{R}^2 \mapsto \mathbb{R}$ $$\zeta_{\varepsilon}(\mathbf{r}) = \frac{1}{\pi^2 \sqrt{\mathbf{r}\cdot \mathbf{r}}} \frac{\varepsilon}{\varepsilon^2 + \mathbf{r}\cdot \mathbf{r}},$$ the integral over $\mathbb{R}^2$ is one independently of $\varepsilon$.
Does the following hold? $$\lim_{\varepsilon \to 0} \zeta_{\varepsilon} = \delta(||\mathbf{r}||)$$
If it does, it must also hold $$\frac{1}{\pi^2} \lim_{\varepsilon \to 0} \eta_{\varepsilon} = ||\mathbf{r}|| \delta(||\mathbf{r}||).$$
Or is this reasoning flawed?
No, this does not converge, take for example $f \equiv 1$: $$\int_0^{\infty} \frac{\varepsilon r}{\varepsilon^2 + r^2} \, \mathrm{d}r = \frac{\varepsilon}{2}\log(\varepsilon^2 +r^2)\big|_0^{\infty} = \infty.$$ What you can do, however, take any non-negative $\eta \in L^1(\mathbb{R}^2)$ with $\|\eta\|_1 = 1$. Then $\eta_{\varepsilon}(x) := \frac{1}{\varepsilon^2}\eta(\frac{x}{\varepsilon})$ defines a Dirac sequence.