Showing that $ \frac{x\sin(y)-y\sin(x)}{x^2+y^2}\rightarrow_{(x,y)\to (0,0)}0$ I would like to show that:
$$ \frac{x\sin(y)-y\sin(x)}{x^2+y^2}\rightarrow_{(x,y)\to (0,0)}0$$
$$ \left| \frac{x\sin(y)-y\sin(x)}{x^2+y^2} \right| \leq \frac{2\vert xy \vert}{x^2+y^2} \leq 1$$
which is not sharp enough, obviously.
How can I efficiently "dominate" the quantity $ \vert x\sin(y)-y\sin(x)\vert$ ?
 A: If $xy=0$ then your function is zero. You could just write for $x,y \neq 0$
$$\left| \frac{x\sin(y)-y\sin(x)}{x^2+y^2} \right| = \frac{|xy|}{x^2+y^2}\left|\frac{\sin x}{x}-\frac{\sin y}{y}\right|  \leq \frac{1}{2}\left|\frac{\sin x}{x}-\frac{\sin y}{y}\right|$$ which tends to zero as $x,y \to 0$.
A: We have by the fundamental theorem of analysis that 
$$x\sin y=x\int_0^y\cos tdt=x\left(y\cos y-\int_0^y-t\sin tdt\right)=xy\cos y+x\int_0^yt\sin tdt$$
and similarly 
$$y\sin x=xy\cos x+y\int_0^xt\sin tdt$$ 
so 
$$|x\sin y-y\sin x|\leq |xy|\cdot|\cos x-\cos y|+|x|\frac{y^2}2+|y|\frac{x^2}2$$
and finally 
$$\frac{x\sin y-y\sin x}{x^2+y^2}\leq \frac 12|\cos x-\cos y|+|x|+|y|,$$
which gives the result.
A: Using polar coordinates:
$\lim_{r->0^+}\frac{r\cos(\theta)\sin(r\sin(\theta))-r\sin(\theta)\sin(r\cos(\theta))}{r^2}=\lim_{r->0^+}\frac{\cos(\theta)\sin(r\sin(\theta))-\sin(\theta)\sin(r\cos(\theta))}{r}=$
(using  De L'hospital rule)
$\lim_{r->0^+}{\cos(\theta)\sin(\theta)\cos(r\sin(\theta))-\sin(\theta)\cos(\theta)\cos(r\cos(\theta))}=$
$\sin(\theta)\cos(\theta)\lim_{r->0^+}\cos(r\sin(\theta))-cos(r\cos(\theta))=0$
