How do I show that this Limit of 2 variables is zero? How do I show that :$$\lim_{(x,y)\to(0,0)}xy\frac{x^2-y^2}{x^2+y^2}=0?$$
I'm stumped...
 A: $$
0 \le (x-y)^2 = x^2+y^2 - 2xy\implies|xy| \le \frac{x^2+y^2}{2}
$$
A: $$0\leq|xy|\frac{|x^2-y^2|}{x^2+y^2}\leq |xy|\to0$$
A: For the sake of some variety in answers, let's see how this goes with the polar substitution $x  = r\cos(\theta)$ and $y  = r\sin(\theta)$ note:
$$ xy \frac{x^2-y^2}{x^2+y^2} =  r^4 \frac{ \sin(\theta)\cos(\theta)(\cos^2(\theta)-\sin^2(\theta))}{r^2} = r^2 h(\theta)$$
Clearly $h(\theta)$ is bounded as $r \rightarrow 0$ hence
$$ \lim_{(x,y) \rightarrow (0,0)} xy \frac{x^2-y^2}{x^2+y^2} = \lim_{r \rightarrow 0} r^2h(\theta) = 0. $$
Of course, the method suggested by the other answers is more rigorous, but in the event the limit does not exist this method sometimes reveals how to obtain inconsistent path limits.
A: Personally, I would take a change of coordinates and look at the limit in polar coordinates.
Substituting $x = r\cos\theta$ and $y = r\sin\theta$, we get
\begin{array}
xxy \frac{x^2-y^2}{x^2+y^2} &=& r^2\cos\theta\sin\theta\frac{r^2\cos^2\theta-r^2\sin^2\theta}{r^2\cos^2\theta + r^2\sin^2\theta} \\ \\
&\equiv& r^2(\cos^3\theta\sin\theta - \cos\theta\sin^3\theta) \\ \\
&\equiv& \frac{r^2}{4}\sin 4\theta
\end{array}
As $r \to 0$, this tends towards 0 for all values of $\theta$.
A: Remember that $\lim_{(x,y)\to (0,0)}f(x,y)=0$ if,  and only if, $\lim_{(x,y)\to (0,0)}|f(x,y)|=0$. Note that, $|x^2-y^2 |\leq | x^2+y^2|$ implies
$$
0
\leq
\left| xy\frac{x^2-y^2}{x^2+y^2} \right|
\leq 
|x||y|\cdot\left| \frac{x^2+y^2}{x^2+y^2}\right|
\leq
|x||y|
$$
By the sandwich theorem the inequalities
$$
0
\leq 
\left| xy\frac{x^2-y^2}{x^2+y^2} \right|
\leq
|xy|
$$
and $
\lim_{(x,y)\to (0,0)}0 =0 $,$
\lim_{(x,y)\to (0,0)}|xy|=0
$
implies 
$$
\lim_{(x,y)\to (0,0)}\left| xy\frac{x^2-y^2}{x^2+y^2} \right|=0
$$
