Limits of Functions with Two Variables: $\lim_{(x,y)\to(0,0)} \frac{xy(x-y)}{(x^2+y^2)}$ $$\lim_{(x,y)\to(0,0)} \frac{xy(x-y)}{(x^2+y^2)}$$ I know that this should be $0$, I'm just not entirely sure how to prove this. I have worked out at least this much:
For all $\epsilon > 0$, $\frac{xy(x-y)}{(x^2+y^2)} <\epsilon $ whenever $0<\sqrt{x^2 +y^2}< \delta$ for all $\delta >0$.
This is the same as to say: $x^2 +y^2 < \delta^2$. Now since this matches the bottom of the fraction, I can do:
$xy(x-y) < \epsilon \delta^2 < \epsilon$ when $\delta \leq 1$.
I feel as though I may have gotten on the wrong track toward the end here. Could someone help guide me to how I can choose my $\delta$ value?
 A: Let's try rewriting in polar form.
$\lim_{(x,y)\to(0,0)} \frac{xy(x-y)}{(x^2+y^2)}$
Now, by using standard polar substitutions for $x=r\cos(\theta)$ and $y=r\sin(\theta)$
$\lim_{r\to0} \frac{(r\cos\theta)^2(r\sin\theta)-(r\sin\theta)^2(r\cos\theta)}{r^2}$
$\lim_{r\to0} \frac{r^3\cos^2\theta\sin\theta-r^3\sin^2\theta\cos\theta}{r^2}$
$\lim_{r\to0} r(\cos^2(\theta)\sin(\theta)-\sin^2(\theta)\cos(\theta))$
The $\theta$ value is insignificant, because it does not matter from which angle we approach the origin, so long as we approach it; namely $r$ must approach $0$. [With the assumption of a real-valued function] 
Consequently, you let $r\to0$ and the resulting expression is...$0$!
A: It is well known that
$$ \left|\frac{xy}{x^2+y^2}\right|\leq 1$$
for all $(x,y)\neq (0,0)$. So you have
$$\left|\frac{xy(x-y)}{x^2+y^2}\right|\leq |x-y|\leq |x|+|y|\leq 2\sqrt{x^2+y^2}.$$
So for any $\epsilon>0$, if you define $\delta=\frac{\epsilon}{2}$ then
$$\sqrt{x^2+y^2}<\delta \Rightarrow \left|\frac{xy(x-y)}{x^2+y^2}\right|<\epsilon.$$
A: Using $(x-y)^2 \geq 0$, we have $xy \leq \dfrac{x^2+y^2 }{2}$. Thus: $0 \leq \left|\dfrac{xy(x-y)}{x^2+y^2}\right|\leq \dfrac{|x-y|}{2}$,and using squeeze lemma, the result follows.
A: Hint:
$$|xy(x-y)|<2\left(\sqrt{x^2+y^2}\right)^3.$$
