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!


  • 4
    $\begingroup$ Close. Take it $\sqrt{\epsilon}$. $\endgroup$ – user207710 Mar 22 '15 at 19:53
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    $\begingroup$ 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

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

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


  • $\begingroup$ 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

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