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.


1 Answer 1


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$$|4x^2-5y^2-(-4)|\leqslant\frac{20\varepsilon}{50}+\frac{25\varepsilon}{50}<\varepsilon.$$

  • $\begingroup$ 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? $\endgroup$
    – Us_55
    Jun 14, 2020 at 18:54
  • $\begingroup$ That was my goal indeed. Perhaps that $45$ would have been a better choice. $\endgroup$ Jun 14, 2020 at 19:34

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