I need to find the limit by just using the $\epsilon$ and $\delta$ definition. The function is the following:
$$\lim_{(x,y)\rightarrow(2,-2)} 4x^2 -5y^2 $$
I already know that $\lim \rightarrow -4$ and that if I were to do the proof knowing the limit, it would be:
If $0 < \sqrt{(x-2)^2 +(y+2)^2} < \delta \Rightarrow |3x^2-4y^2+4|<\epsilon $
Can the limit be proved to exist even if I don't know the exact value? If so, how can I do it? Given the result of the limit, how can I manipulate the expression: $|3x^2-4y^2+4|$ so that I can give the $\epsilon$ and $\delta$ relationship? I read that I can use polar coordinates but I would like to know if I can do it without them.
Thanks for the help.
The map $(x,y)\mapsto4x^2-5y^2$ looks continuous, right?! So, it is to be expected that the limit at $(2,-2)$ is $4\times2^2-5\times(-2)^2=-4$.
Note that\begin{align}4x^2-5y^2-(-4)&=4x^2-16-(5y^2-20)\\&=4(x-2)(x+2)-5(y-2)(y+2).\end{align}So, if $\|(x,y)-(2,-2)\|<1$, then both $|x-2|$ and $|y+2|$ are smaller than $1$ and therefore $|x+2|$ and $|y-2|$ are smaller than $5$. So,\begin{align}|4x^2-5y^2-(-4)|&=|4(x-2)(x+2)-5(y-2)(y+2)|\\&\leqslant20|x-2|+25|y+2|\end{align}So, given $\varepsilon>0$, take $\delta=\min\left\{1,\frac\varepsilon{50}\right\}$. If $\|(x,y)\|<\delta$, then$$|4x^2-5y^2-(-4)|\leqslant\frac{20\varepsilon}{50}+\frac{25\varepsilon}{50}<\varepsilon.$$