Use the epsilon/delta definition to find and prove the limit:

$\displaystyle \lim_{\textbf{X}\to\textbf{X}_0}f(\textbf{X})=\fra c{x^3-y^3}{x-y}$, $\displaystyle \textbf{X}_0=(1,1)$.My notes so far:

We have $\displaystyle f(\textbf{X})=\frac{x^3-y^3}{x-y}=x^2+xy+y^2,x\neq y$. So

$\displaystyle |f(\textbf{X})-f(\textbf{X}_0)|=|x^2+xy+y^2-x_0^2-x_0y_0-y_0^2|$

$\displaystyle =|(x-x_0)(x+x_0)+(y-y_0)(y+y_0)+xy-x_0y_0|$

$\displaystyle \leq|(x-x_0)(x+x_0)|+|(y-y_0)(y+y_0)|+|xy-x_0y_0|$

$\displaystyle \leq|xy-x_0y_0|+|\textbf{X}-\textbf{X}_0|(|x+x_0|+|y+y_0|)$.

That last inequality is because

$\displaystyle |\textbf{X}-\textbf{X}_0|=|(x-x_0,y-y_0)|=\sqrt{(x-x_0)^2+(y-y_0)^2}$

$\displaystyle \geq|x-x_0|,|y-y_0|$.

But I don't seem to be getting anywhere with that. I've also looked at the Cauchy-Schwarz inequality, but that doesn't offer me any obvious (to me) help.

Any ideas? Tricks? A complete solution? This isn't supposed to be a difficult problem, but it's the first of its kind which I have attempted!

Thanks in advance!

EDIT: I solved it on my own. Let $\displaystyle \delta=-1+\sqrt{1+\frac{\epsilon}{3}}>0$