1
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.