# Epsilon delta definition of limits with 2 variables and no E or D values

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.

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

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

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

• 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? – mathjohnn Jan 12 at 18:44
• Nevermind, I got it. Thank you! – mathjohnn Jan 12 at 18:50
• 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? – Peter Szilas Jan 12 at 19:04
• mathjohnn.Edited the last line. – Peter Szilas Jan 12 at 19:38

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.

• Thank you for your help! – mathjohnn Jan 12 at 18:51