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## Homework Statement

Find the limit [tex]\lim_{(x,y)\to(2,2)}\frac{x^3-y^3}{x-y}[/tex]

## Homework Equations

[itex]\epsilon - \delta[/itex], baby:

If the limit L exists,

[tex]

\forall \: \epsilon \: \exists \: \delta: 0 < \sqrt{(x-a)^2+(y-b)^2} < \delta \rightarrow |f(x,y)-L| < \epsilon

[/tex]

## The Attempt at a Solution

By approaching the point (2,2) along the lines x=2 and y=2 I managed to figure out that the limit is likely 12. Now I just need to prove it, and I'm stuck.

Now

[tex]0 < \sqrt{(x-2)^2+(y-2)^2} < \delta \leftrightarrow (x-2)^2+(y-2)^2 < \delta^2[/tex]

and using long division

[tex]x^3-y^3 = (x-y)(x^2+xy+y^2) \rightarrow \frac{x^3-y^3}{x-y}=(x^2+xy+y^2)[/tex]

I tried deriving the expression for delta from epsilon like this:

[tex]

\begin{split}

|(x^2+xy+y^2)-12| &= |x^2-4x+4+y^2-4x+4+xy+4x+4y-20|\\

&\leq |x^2-4x+4+y^2-4x+4|+|xy+4x+4y-20|\\

&\leq \delta^2 + |xy+4x+4y-20|

\end{split}

[/tex]

Almost there, except I'm stuck with that pesky |xy+4x+4y-20|, that I don't know how to get rid of. Am I taking the wrong approach to this entirely? Help, I'm desperate.