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

Question

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.

Answer 1

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

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

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

Answer 2

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.