<s>[INST] Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists. 
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
  exists.

Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x \to 0} \frac{x^3+x^3}{x^2+x^2} = x = 0$ 
is this method correct?
 [/INST] A better way to solve this problem is to use polar coordinates
$$x=r\cos \phi$$
$$y=r\sin \phi$$
$$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}=\lim_{r\to 0}\frac{r^3(\cos^3 \phi+\sin^3 \phi)}{r^2}=\lim_{r\to 0}\left[r(\cos^3 \phi+\sin^3 \phi)\right].$$
Because $|\cos^3 \phi + \sin^3\phi|\leq 1$ the limit exists (by the sandwich theorem) and is zero. 
Remark: If you assume that the limit exists then you could also choose $y=0$ and let $x \to 0$ (because it does not matter how you approach $(0,0)$).
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