1. The problem statement, all variables and given/known data
Hi, I got this other limit here, which I think I've solved rightly. I've used polar coordinates, and I've found a limit that doesn't depends on theta, but I've got some doubts, I've been reading some old threads like https://www.physicsforums.com/showthread.php?t=114567&page=2" (there a limit likely the one I'm threatening here but not exactly the same when they talk about the definition of the neighbors at the point), and since I've looked at many trajectories and everything indicates the solution its fine, I'm not sure about if I'm missing something.

I am okay to here but how did you get the next term?

And, of course, [tex]\frac{2r}{r^4}= -r^{-3}[/tex], which has no limit as r goes to 0, not "[tex]-r^3[/tex]".

Yes, but your calculation is not valid!

From [tex]\lim_{r\to 0^+}\frac{ln(r^2)}{\frac{1}{r^2}}=\lim_{r\to 0^+} \frac{2ln(r)}{r^{-2}}[/tex] you can use L'Hopital's rule to get [tex]\lim_{r\to 0^+}\frac{2\frac{1}{r}}{-2r^{-3}}= \lim_{r\to 0^+}-r^2= 0[/tex].

I've tried to solve the part on the right using polar coordinates, and I get to

[tex]\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}=\lim_{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2\cos^2\theta+r^4\sin^4\theta}}=\lim_{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2(\cos^2\theta+r^2\sin^4\theta)}}=\lim_{r \to{0}+}{\displaystyle\frac{\sin^2\theta}{\cos^2\theta+r^2\sin^4\theta}}=\displaystyle\frac{\sin^2\theta}{cos^2\theta}[/tex]
Which means this limit doesn't exists, but as it is multiplied by something that tends to zero, the limit should be zero. Is this right?