$f(x,y)=\frac{x^3+y^3}{x-y}, \; \;(x,y)\neq(0,0)$, otherwise $0$
If I use $x=r\cos(t), \; y=r\sin(t)$ & $r\to 0$, I get the limit equal to $0$. Is this sufficient to prove that the limit of the function is $0$ and that it is continuous?
Basically, I want to know that, if I apply the polar coordinates and if I get a finite constant limit, say $k$ (such a limit will not depend on $t$), then can I claim that the limit of the functions is equal to $k$? -------- (1)
According to my understanding, we are allowing $t$ to vary anywhere between $0 $ to $2\pi$ while taking the limit of $r$ to $0$. So this must mean that we are covering all possible tracks/paths in reaching the point $(0,0)$. Does this interpretation make any sense?
EDIT:- Ok I realise polar coordinates don't work since
$\lim_{r\to 0} \frac{r^3(\cos^3(t)+\sin^3(t))}{r(\cos(t)-sin(t))} = \lim_{r\to 0}\frac{r^2(\cos^3(t)+\sin^3(t))}{(\cos(t)-sin(t))}$ will give problem when $t=\frac\pi4$
But can I still get the answer to $(1)$?
I don't understand how you were able to get the limit $0$ with polar coordinates. The function $f$ has no limit at $(0,0)$. Just note that$$(\forall n\in\mathbb{N}):f\left(\frac1n+\frac1{n^3},\frac1n\right)=2+\frac{3}{n^2}+\frac{3}{n^4}+\frac{1}{n^6}.$$