What is $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2} $ Find the following limit 
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2}$$

This question has been bugging me for some time. Couldn't find it anywhere on the Internet
You can cancel a common factor:
$$\frac{x^3-y^3}{x^2-y^2}=\frac{(x-y)(x^2+xy+y^2)}{(x-y)(x+y)}=\frac{x^2+xy+y^2}{x+y} $$
(This also continuously extends the function to most of the line $x=y$).

Along any straight line $(x,y)=(at,bt)$, $t\to 0$, this is $\frac{a^2 +ab+b^2}{a+b}\cdot t$ and tends to $0$. But this of course does not work for the case $b=-a$ where the expression is undefined.
However, convergence along straight lines is not convergence!
If we consider $(x,y)=(t+t^2,-t)$, $t\to 0$, we arrive at
$$\frac{(t+t^2)^2+(t+t^2)(-t)+(-t)^2}{t+t^2-t} =\frac{t^2+t^3+t^4}{t^2}=1+t+t^2\to 1$$
i think the limit is 0
switch to polar coordinates and you will see it. you can also simplify the equation: x^3 - y^3 = (x-y)(x^2+xy +y^2)
good luck