For any $\epsilon > 0$, I want to show that $|2x^2 -3y^2-2|< \epsilon$ whenever
$0<\sqrt{(x-5)^2 + (y-4)^2}< \delta$.
What I've tried so far is expanding out the argument in the absolute value to get:
$|2x^2-20x +20x +50 – 50 -3y^2 +24y -24y + 48 -48-2|$ so that I can get
$|2(x-5)^2 -3(y-4)^2 +20x -24y -100|$.
I could use the triangle inequality to get that
$|2(x-5)^2 -3(y-4)^2 +20x -24y -100| \le |2(x-5)^2 -3(y-4)^2| +|20x -24y -100|$ and use the triangle inequality again to get
$|2(x-5)^2 -3(y-4)^2| +|20x -24y -100| \le |2(x-5)^2| + |3(y-4)^2| +|20x -24y -100|$ so that I could possibly do
$|2(x-5)^2| + |3(y-4)^2| +|20x -24y -100|< 2\delta^2 + 3\delta^2 + |20x -24y -100|$
But I'm not sure how to proceed from there. It would be helpful if someone could possibly point out a better method? That, or help with possible next steps?
Best Answer
Make $\delta <1$,
\begin{align} |2x^2-3y^2-2| &= |2(x-5+5)^2-3(y-4+4)^2-2|\\ &=|2[(x-5)^2+10(x-5)+25]-3[(y-4)^2+8(y-4)+16]-2|\\ &=|2(x-5)^2+20(x-5)-3(y-4)^2-24(y-4)| \\ &\le 2\delta^2 + 20 \delta +3\delta^2+24 \delta \\ &\le 49 \delta \end{align}