3
$\begingroup$

I am wondering about a beginner proof I am trying to do, and also a more general question.

I am working on the question, $$\lim_{(x,y) \to (0,0)}\frac{x^2y^2}{x^2+y^2}$$

So I looked at a few cases and it seemed to me that it might be reasonable that the limit is 0, so I tried to do an epsilon delta proof.

that is, I am under the impression that if I can show that for any $\epsilon >0$ there exists a $\delta > 0$, such that if $0< \sqrt{x^2+y^2}< \delta $ then $| \frac{x^2y^2}{x^2+y^2}-0|< \epsilon$ ( also I can remove the absolute value signs because we have only squares).

So then I thought I could do something like, $$x^2 \le x^2+y^2 \rightarrow \frac{x^2}{x^2+y^2} \le 1 \rightarrow \frac{x^2y^2}{x^2+y^2} \le y^2 \le x^2+y^2$$ ( using that $y^2 \le x^2+y^2$ as well)

But now I am not sure how to proceed, should I say let $\delta=\epsilon$ ? Is that even valid to say. Also, I have not looked at solutions or anything so I am not sure if my work is correct either, so i appreciate any comments/answers!

Thankyou.

Improve this question
$\endgroup$
  • 4
    $\begingroup$ Close. Take it $\sqrt{\epsilon}$. $\endgroup$ – user207710 Mar 22 '15 at 19:53
  • 1
    $\begingroup$ Yes, $\delta$ is allowed to depend on $\epsilon$. $\endgroup$ – shalop Mar 22 '15 at 19:59
  • $\begingroup$ @Shalop Yea I understand that, but I was more asking can it actually = directly? i.e. $\delta = \epsilon$ ? $\endgroup$ – Quality Mar 22 '15 at 20:06
  • $\begingroup$ Consider $\lim_{x\to 0} x=0$. Then $\delta =\epsilon$ suffices, since $0<|x-0|<\delta\implies 0<|x-0|<\epsilon$. Your $\delta$ can be any function of $\epsilon$, given it actually implies the function being within $\epsilon$ of the limit. $\endgroup$ – user26486 Mar 22 '15 at 20:08
2
$\begingroup$

Let $\epsilon > 0$. Take $\delta = \sqrt{\epsilon}$. Let $(x,y) \in \mathbb R^2$:

$$||(x,y) - (0,0)||_2 < \delta \implies \sqrt{x^2 + y^2} <\delta \implies x^2 + y^2 < \delta^2 \implies x^2 + y^2 < \epsilon $$

etc.

Improve this answer
$\endgroup$
  • $\begingroup$ Thanks, what is the little 2 beside the norm by the way? $\endgroup$ – Quality Mar 22 '15 at 20:06
  • $\begingroup$ My pleasure. It is to signify that this is the Euclidean norm. There are other types of norms: en.wikipedia.org/wiki/Norm_%28mathematics%29 $\endgroup$ – user207710 Mar 22 '15 at 20:10

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.