The relativistic heat equation or telegraphers equation is: $$ (\alpha\partial_t^2 + \beta\partial_t - \omega\,\nabla^2_{\text{3D}})G_R = \delta $$
if $\alpha \rightarrow 0$ the solution must converge to the non-relativistic heat equation $$ (\beta\partial_t - \omega\,\nabla^2_{\text{3D}})G = \delta $$
From Wikipedia https://en.m.wikipedia.org/wiki/Green%27s_function#Table_of_Green%27s_functions the green function for the 3D relativistic heat equation with $\gamma = \frac{\beta}{2\alpha}$ and $c = \sqrt{\frac{\omega}{\alpha}}$ is: $$ (\partial_t^2 + \frac{\beta}{\alpha}\partial_t - \frac{\omega}{\alpha}\,\nabla^2_{\text{3D}})G_R = \frac{1}{\alpha}\delta $$ $$ (\partial_t^2 + 2\gamma\partial_t - c^2\,\nabla^2_{\text{3D}}) G'_R = \delta $$ so $G_R = \frac{G'_R}{\alpha}$ $$ G'_R =\frac{e^{-\gamma t}}{20\pi} \left[\left(8-3e^{-\gamma t}+2\gamma t+4\gamma^2t^2\right)\frac{\delta(ct-r)}{r^2}+\frac{\gamma^2}{c}\Theta(ct - r)\left(\frac{1}{cu}I_1\left(\frac{\gamma u}{c}\right)+\frac{4 t}{u^2}I_2\left(\frac{\gamma u}{c}\right)\right)\right] $$
with $ u=\sqrt{c^2t^2-r^2}$
if $\alpha \rightarrow 0$ then $\gamma, c \rightarrow \infty$ so: $$ G'_R =\frac{e^{-\gamma t}}{20\pi} \left[\frac{\gamma^2}{c}\Theta(t)\left(\frac{1}{cu}I_1\left(\frac{\gamma c t}{c}\right)+\frac{4 t}{c^2 t ^2}I_2\left(\frac{\gamma c t}{c}\right)\right)\right] = \frac{e^{-\gamma t}}{20\pi} \left[\frac{\gamma^2}{c}\Theta(t)\left(\frac{1}{c^2 t}I_1\left(\frac{\gamma c t}{c}\right)+\frac{4 t}{c^2 t ^2}I_2\left(\frac{\gamma u}{c}\right)\right)\right] $$ The modified bessle function according to https://personal.math.ubc.ca/~cbm/aands/abramowitz_and_stegun.pdf is asymptotic to $I_\nu(z) = \frac{e^z}{\sqrt{2\pi z}}$ $$ G'_R =\frac{\exp{(-\gamma t)}}{20\pi} 5 \left[\frac{\gamma^2}{c^3 t}\Theta(t)\left(\frac{exp(\gamma t))}{\sqrt{2\pi \gamma t}})\right)\right] exp(-\frac{\gamma r^2}{2 c^2 t})=\frac{1}{4\pi} \left[\frac{\gamma^2}{c^3 t}\frac{1}{\sqrt{2\pi \gamma t}}\right]\Theta(t)exp(-\frac{\gamma r^2}{2 c^2 t}) $$
Now substituting in the equation $\alpha$, $\beta$, $\omega$ we get: $$ G_R = \frac{G'_R}{\alpha} = \frac{1}{4\pi \alpha} \left[\frac{(\frac{\beta}{2\alpha})^2}{(\frac{\omega}{\alpha})^\frac{3}{2} t}\frac{1}{\sqrt{2\pi \frac{\beta}{2\alpha} t}}\right]\Theta(t) exp(-\frac{\beta r^2}{4 \omega t}) = \frac{1}{4\pi} \left[\frac{(\frac{\beta}{2})^2}{(\omega)^\frac{3}{2} t}\frac{1}{\sqrt{2\pi \frac{\beta}{2} t}}\right]\Theta(t) \frac{\alpha^\frac{3}{2}\alpha^\frac{1}{2}}{\alpha^3}exp(-\frac{\beta r^2}{4 \omega t}) $$ Therefore:
$$ G_R = \frac{1}{4\pi} \left[\frac{(\frac{\beta}{2})^2}{(\omega)^\frac{3}{2} t}\frac{1}{\sqrt{2\pi \frac{\beta}{2} t}}\right]exp(-\frac{\beta r^2}{4 \omega t})\Theta(t) \frac{1}{\alpha} = \frac{\beta^\frac{3}{2}}{16(\omega \pi t)^\frac{3}{2}}exp(-\frac{\beta r^2}{4 \omega t})\Theta(t) \frac{1}{\alpha} $$ the solution is approximately correct to the non-relativistic one in the dependence of time and space but there are issues with the constants and it must not depend on $\alpha$ the non-relativistic approximation.
The non-relativistic green function of the heat equation is: $$ G = \frac{\beta^\frac{1}{2}}{(4 \pi \omega t)^{\frac{3}{2}}}exp(-\frac{r^2 \beta}{4 \omega t}) $$
I do not understand if there is something wrong with the solution found on Wikipedia or I am doing some errors in the approximations