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I have seen plenty of epsilon delta examples, but am not sure how to apply them to this problem. The question states "Using the $\epsilon$ − δ definition of limits, show that $\lim\limits_{x, y \to (0,0)} xy\frac{x^2-y^2}{x^2+y^2}=0$. I know how to prove a limit exists by showing delta > epsilon, but nothing at this level. Any help is appreciated, thank you.

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$(x,y)\not =(0,0)$

Let $\epsilon >0$ be given.

$f(x,y)=:|xy|\dfrac{|x^2-y^2|}{x^2+y^2}\le$

$|xy|\dfrac{x^2+y^2}{x^2+y^2}= |xy|\lt (x^2+y^2)$.

Choose $\delta = √\epsilon$.

Then

$\sqrt{x^2+y^2} < \delta$ implies

$f(x,y) < x^2+y^2 < \delta^2 < \epsilon$.

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  • $\begingroup$ Thank you for answering. I understand that the f(x) is within $\epsilon$ , but how does that prove that the limit is equal to 0? Or does it just prove a limit exists, and is implied to be 0? $\endgroup$ – mathjohnn Jan 12 at 18:44
  • $\begingroup$ Nevermind, I got it. Thank you! $\endgroup$ – mathjohnn Jan 12 at 18:50
  • $\begingroup$ mathjohnn.It is $f(x,y)$, edited.Now the limit for 1 variable: $\lim_{x \rightarrow 0}f(x)=L$ if for every $\epsilon >0$ there is a $\delta$ s.t. $|x-0|<\delta$ implies $|f(x)-L|<\epsilon$. For 2 variables we have: $\sqrt{x^2+y^2} <\delta$(distance from 0) implies $|f(x,y)-L|<\epsilon$.$\epsilon$ given, arbitrarily small, find a $\delta$.OK? $\endgroup$ – Peter Szilas Jan 12 at 19:04
  • $\begingroup$ mathjohnn.Edited the last line. $\endgroup$ – Peter Szilas Jan 12 at 19:38
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We need to show that for all $\epsilon>0$ there exists a $\delta>0$ such that if $|x|,|y|<\delta$ then $$ \left|xy\frac{x^2-y^2}{x^2+y^2}\right|<\epsilon. $$ I claim that $\delta=\sqrt{\epsilon}$ works. Indeed, since $|x^2-y^2|\leq x^2+y^2$ we have the inequality $$ \left|xy\frac{x^2-y^2}{x^2+y^2}\right|\leq |x||y|< \delta^2, $$ so if $\delta=\sqrt{\epsilon}$ we get the desired bound.

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  • $\begingroup$ Thank you for your help! $\endgroup$ – mathjohnn Jan 12 at 18:51

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