# How can I find a limit just using $\epsilon$ and $\delta$

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$$|4x^2-5y^2-(-4)|\leqslant\frac{20\varepsilon}{50}+\frac{25\varepsilon}{50}<\varepsilon.$$
• Thanks for your time and for your help. I just have one question, from where do you take the $\epsilon / 50$? Is it a denominator that makes sense when doing the addition in the last part? – Us_55 Jun 14 at 18:54
• That was my goal indeed. Perhaps that $45$ would have been a better choice. – José Carlos Santos Jun 14 at 19:34