# Multivariable Limit: how to prove it exists?

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!

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,$$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$

• Sorry for the long delay, but thank you very much for the answer! – Miguel Ferreira Mar 17 at 11:14

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 $$\left|\frac{x^2+y^2}{\log(x^2+y^2)}-0\right|\leq \left|x^2+y^2 \right|\to 0$$

• Sorry for the long delay! Thank you for the answer!:) – Miguel Ferreira Mar 17 at 11:14