Prove that the limit of $3x^{2} + xy - 2y^{2} = 0$ as $(x,y) \rightarrow (2,3)$ using $\epsilon - \delta$ definition.

Question

This is what I've done so far:

$|3x^2 + xy - 2y^2| < \epsilon$ whenever $\sqrt{(x-2)^2 + (y-3)^2} < \delta.$

I know that,

$|3x^2 + xy + 2y^2| = |3x - 2y||x+y|$

$=|3x +6 -2y - 6||x+y|$

$= (3|x-2| - 2|y-3|)*|x+y|$

and

$|x-2| = \sqrt{(x-2)^2} \leq \sqrt{(x-2)^2 + (y-3)^2} < \delta.$ Similarly,

$|y-3| = \sqrt{(y-3)^2} \leq \sqrt{(x-2)^2 + (y-3)^2} < \delta.$

Now I do not know what to do with $|x+y|$. Hope someone can help me. I am new to proofs of these.

Answer 1

You can take $\delta=\min\{\epsilon,1\}$ to control both $x$ and $y$ not too large around $(2,3)$: For $\sqrt{(x-2)^{2}+(y-3)^{2}}<\delta\leq 1$, then $|x-2|\leq\sqrt{(x-2)^{2}+(y-3)^{2}}$ and hence $|x-2|<1$, and $|x|\leq|x-2|+2<3$. Similarly, $|y|\leq|y-3|+3<4$ so $|x+y|\leq|x|+|y|<7.$