<s>[INST] 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.
 [/INST] 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.
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