What is $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$? Given is the function $f(x,y)=\frac{x^3-y^3}{x^2-y^2}$ defined for $(x,y)\in\mathbb{R^2}, \, x^2\neq y^2$
I need to find the limit of $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$. 
I have tried a few paths, $(x,mx),\, (x,mx^2), \, (x,sin(x)), \, (x,tan(x))$ but all of them yield the same result, $0$. 
So I tried to prove the limit exists using the epsilon-delta definition. So:
$$\left|\frac{x^3-y^3}{x^2-y^2} - 0\right| = \left|\frac{(x-y)(x^2+y^2+xy)}{(x-y)(x+y)}\right|= \frac{x^2+y^2+xy}{\left|x+y\right|}<\epsilon$$
Here I am stuck and do not know how to proceed.
 A: The problem is that there is no limit, only pathwise limits (which of course depend on the path to zero).
To see this one could rewrite the expression:
$${x^3-y^3\over x^2-y^2} = {(x-y)(x^2+xy+y^2)\over(x-y)(x+y)} = {(x+y)x - y^2\over (x+y)} = x - {y^2\over x+y}$$
Now of course $x\to 0$, but the term $y^2/(x+y)$ can be selected arbitrarily by having $x = y^2/C - y$. So along the parabola $x = y^2/C - y$ aproaching $0$ we have the limit of the expression being $-C$.  
But this can't happen if the limit actually exists.
A: A method is to consider $y=mx$
Note : $m=m(x,y)$ is variable, it is the ratio of converging rates of $y$ and $x$.

$$f(x,y)=\frac{x^2(1+m^2+m)}{x(1+m)}=x\frac{1+m+m^2}{1+m}$$
If $m\to\pm\infty\quad f(x,y)\sim\frac{m^2}{m}x=mx=y\to 0$
If $m\to0$ then $f(x,y)\sim x\to 0$
If $m$ is bounded let's have $u=m+1\quad$ ($u$ also variable).
$\displaystyle{f(x,y)=x\frac{u+(u-1)^2}{u}=x\frac{u^2-u+1}{u}=x(u-1+\frac1u)=xm+\frac xu\sim \frac xu}$ 
because if $m$ is bounded then $mx\to 0$.
And that is a big problem : for $u=kx$ with $k\in\mathbb R$ we have $f(x,y)\to\frac1k$ and $m=kx-1$ is effectively bounded, thus a valid case.
Conclusion :
There is no limit in $(0,0)$ because for $k\in\mathbb R,\ x\to 0,\ f(x,kx^2-x)\to\frac1k$
$f$ has multiple limits depending on the path chosen to go to $(0,0)$.
