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I've recently started Calculus II and I'm not being able to understand how to prove if a limit does exist, or not.

Having $f(x,y) = \frac{x^2+y^2}{ln(x^2+y^2)},$ if $x^2+y^2<1 \wedge (x,y)\neq (0,0),$

and $f(x,y) = 0$ if $(x,y) = (0,0),$

how can I study the continuity of the function at the origin?

I've already done this: 1) $\lim_{x\to 0} f(x,0)$ and 2) $\lim_{y\to 0} f(0,y)$ , and they are both equal.

Then I've solved the following limits: $\lim_{x\to 0} f(x,mx)$ and $\lim_{x\to 0} f(x,kx^2)$, and I think they are both 0; is this enough to prove the continuity of the function at (x,y) = (0,0)?

And one more thing: how can I prove if the limit exists by the definition?

Thank you very much!

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Use polar coordinates: if $(x,y)=r(\cos\theta,\sin\theta)$, then$$f(x,y)=\frac{r^2}{\log(r^2)}=\frac{r^2}{2\log r}$$and$$\lim_{r\to0}\frac{r^2}{2\log r}=0.$$Therefore,$$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$

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  • $\begingroup$ Sorry for the long delay, but thank you very much for the answer! $\endgroup$ – Miguel Ferreira Mar 17 at 11:14
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In this case you can just change variables... Setting $u=x^2+y^2$, the limit can be computed as $$ \lim_{u\to 0^+} \frac{u}{\log u} = 0. $$ Regarding your first question, it is not enough to compute limits along straight lines and parabolas to show that a limit exists. Also, you can see that for small enough $x^2+y^2$ you have that $|\log(x^2+y^2)|> 1$ and so $$ \left|\frac{x^2+y^2}{\log(x^2+y^2)}-0\right|\leq \left|x^2+y^2 \right|\to 0 $$

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    $\begingroup$ Sorry for the long delay! Thank you for the answer!:) $\endgroup$ – Miguel Ferreira Mar 17 at 11:14

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