If I am solving a limit on 2 variables and after changing to polar coordinates the angle dissapears, what does that mean?

Question

For example, I have: $\lim_{(x,y)\to(0,0)}\frac{e^{x^2+y^2}\ln(2(x^2+y^2)+1)-\frac{3}{2}(x^2+y^2)-1}{x^2+y^2}$

After a change to polar coordinates: $\lim_{r\to0}\frac{e^{r^2}\ln(2(r^2)+1)-\frac{3}{2}(r^2)-1}{r^2}$

I think that since the angle disappears the function has some sort of radial symmetry, so it doesn't matter how do I approach to $(0,0)$. After that I solved the limit like I would solve a single variable limit using L'Hopital's and the limit was equal to $\frac{3}{2}$ which is correct. Am I correct in assume that every time the angle disappears I can evaluate the limit as if it were one variable? Or what happened there? (Sorry if I have some spelling mistakes, I'm not english native speaker).

Answer 1

Yes, every time the variables show up they show up in the form of $r^2$ as your transformation shows, so in polar the angle is indeed not needed and you get a one variable limit

Answer 2

It means that the limit exists, if the limits depends on the angle in (0,0) then your function is indeterminate in (0,0) and the limit doesn't exists, it happens because your function is symmetrical.