Disprove or prove using delta-epsilon definition of limit that $\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$ I want to prove if the following limit exists, using epsilon-delta definition, or prove it doesn't exist:$$\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$$
My attempt: 
First I proved some directional limits, like for $y=mx$ , and $y=ax^n$, and for all of them I got 0. So I conjectured that this limit exists and it's 0. Then I have to prove:
$$\forall \delta \gt 0 : \exists \epsilon \gt 0 : \|(x,y)\| \lt \epsilon \rightarrow \left| \frac{x^3-y^3}{x^2-y^2}\right| \lt \delta$$
First I noted that $\frac{x^3-y^3}{x^2-y^2} = \frac{(x^2+xy+y^2)(x-y)}{(x+y)(x-y)} = \frac{x^2+xy+y^2}{x+y}$.
Then I did $\left|\frac{x^2+xy+y^2}{x+y}\right| \leq \left|\frac{x(x+y)}{x+y}\right|+\frac{y^2}{\vert x+y\vert} = \vert x \vert + \frac{y^2}{\vert x+y\vert}$
Using $\|(x,y)\| = \vert x\vert + \vert y\vert$ and assuming  $\|(x,y)\| \lt \epsilon$
$\vert x\vert + \vert y\vert  = \vert x \vert + \frac{ y^2}{\vert y \vert} \geq \vert x \vert+\frac{y^2}{\vert y\vert+\vert x\vert}$
but I can't continue from that since $\vert x + y \vert \leq \vert x \vert + \vert y \vert$
.
I don't know what else to try.

Use polar coordinates $x=r\cos\theta$, $y=r\sin\theta$.  Because of the domain we have $\cos\theta\ne\pm\sin\theta$.  Substituting into your formula gives
$$\frac{x^3-y^3}{x^2-y^2}=r\,\frac{\cos^3\theta-\sin^3\theta}{\cos^2\theta-\sin^2\theta}\ .$$
Now it is not hard to prove that
$$\frac{\cos^3\theta-\sin^3\theta}{\cos^2\theta-\sin^2\theta}\to\infty\quad
  \hbox{as}\quad \theta\to\Bigl(\frac{3\pi}{4}\Bigr)^{\textstyle-}\ .$$
So consider the path given by
$$r=\frac{\cos^2\theta-\sin^2\theta}{\cos^3\theta-\sin^3\theta}\ ,\qquad
  \theta\to\Bigl(\frac{3\pi}{4}\Bigr)^{\textstyle-}\ .$$
This will approach the origin, but for every point on this path you have
$$\frac{x^3-y^3}{x^2-y^2}=1\ .$$
So the limit does not exist.
