Limit by polar coordinates $\lim_{(x,y)\to (0,0)} \frac{\sin^2(xy)}{x^2+y^2}=0$ I need to prove the following limit and I would like to have some feedback about my thought process as I'm still a beginner. 
$$\lim_{(x,y)\to (0,0)} \frac{\sin^2(xy)}{x^2+y^2}=0$$
My proof: 
I use polar coordinates and I get an indeterminate form $[\frac{0}{0}]$
$$ \lim_{\rho \to 0} \frac{\sin^2(\rho^2\cos(\theta)\sin(\theta))}{\rho^2} =\lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho} \cdot \lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}$$
Solving 
$$\lim_{\rho \to 0} \frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}\leq\left|\frac{\sin(\rho^2\cos(\theta)\sin(\theta))}{\rho}\right|\leq\frac{\rho^2\lvert\cos(\theta)\rvert\lvert\sin(\theta)\rvert}{\rho}\leq \rho\to 0$$
regardless of $\theta$. Thank you for your help! 
 A: I would write
$$\frac{\sin^2(xy)}{(xy)^2}\times\frac{(xy)^2}{x^2+y^2}$$
and now we use AM-GM inequality
$$x^2+y^2\geq 2|xy|$$
so
$$\frac{(xy)^2}{x^2+y^2}\le \frac{x^2y^2}{2|xy|}=\frac{1}{2}|xy|$$ and this tends to Zero, if $x,y$ are tending to Zero.
A: You may use the fact that
$$
|\sin x|\le |x|
$$
for all $x\in\mathbb R$.
Then, for $(x,y)\ne (0,0)$, 
$$
0\le \frac{\sin^2(xy)}{x^2+y^2}=\frac{\sin^2(r^2\cos\vartheta\sin\vartheta)}{r^2}\le 
\frac{(r^2\cos\vartheta\sin\vartheta)^2}{r^2}=r^2\cos^2\vartheta\sin^2\vartheta\le r^2\to 0
$$
as $(x,y)\to (0,0)$.
A: You are practically done. You have shown
$$0\leq\left\vert\frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}\right\vert\leq \rho$$
and thus, by the squeeze theorem
$$\lim_{\rho\rightarrow 0} \left\vert\frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}\right\vert =0$$
Since the absolute value is continuous, this means 
$$\left\vert\lim_{\rho\rightarrow 0} \frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}\right\vert =0 \Rightarrow \lim_{\rho\rightarrow 0} \frac{\sin(\rho^2\sin\theta\cos\theta)}{\rho}=0$$
and you are done.
