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I'm kind of lost on this problem:

The question says to state whether the limit exists, find its value and prove it:

$\lim_{(x,y) \rightarrow (1,1)} \frac{x^3-y^3}{x^2-y^2}$

So after evaluating several paths (and graphing the contour plot) I know that the limit exists and it's $\frac{3}{2}$. But when it comes to the epsilon delta proof, I'm lost.

I know I have to prove that $\forall \epsilon>0, \exists \delta=\delta(\epsilon) >0$ such that if $\ 0<\left\Vert(x,y) - (1,1)\right\Vert<\epsilon \Rightarrow \left\vert\frac{x^3-y^3}{x^2-y^2} - \frac{3}{2}\right\vert< \epsilon $

I've done other epsilon delta proofs, but the limits were of the form $\lim_{(x,y) \rightarrow (0,0)} f(x,y) = 0$ and now I don't know how to start this one or which inequalities use.

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2 Answers 2

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Berfore we begin, it is worth noting that: $\frac {x^3-y^3}{x^2 - y^2} = \frac {x-y}{x-y}\frac {x^2 + xy + y^2}{x+y}$

$|\frac {x^3 - y^3}{x^2-y^2} - \frac 32| = |\frac {2x^3 - 2y^3 + 3x^2 - 3y^2}{2(x^2-y^2)}|$

let $d((x,y),(a,b)) = \max(|x-a|,|y-b|)$ be our distance metric. Validate that this is a metric if you haven't seen it before.

$|\frac {(x-y)(2(x-1)^2 + 2(x-1)(y-1) + 2(y-1)^2 + 3(x+1) + 3(y+1))}{2(x-y)((x-1)+(y-1) +2)}|$

$|\frac {x^3 - y^3}{x^2-y^2} - \frac 32| < |\frac {6\delta^2 + 6\delta}{4}|$

Let $\delta = \max (1,\frac {\epsilon}{3})$

$|\frac {x^3 - y^3}{x^2-y^2} - \frac 32|<3\delta < \epsilon$

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  • $\begingroup$ The same $\delta$ works for the 'usual metric' on $\mathbb R^{2}$ also. $\endgroup$ Commented Feb 2, 2018 at 6:21
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Epsilon-delta (and "evaluating several paths (and graphing the contour plot)") are really overkill here:$$\frac{x^3-y^3}{x^2-y^2}=\frac {(x-y)f(x,y)}{(x-y)g(x,y)}$$with $f(x,y)=x^2+xy+y^2$ and $g(x,y)=x+y$, so $$\lim_{(x,y)\to(1,1)}\frac{x^3-y^3}{x^2-y^2}=\lim_{(x,y)\to(1,1)}\frac{f(x,y)}{g(x,y)}=\frac{f(1,1)}{g(1,1)}=\frac32.$$

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