Limit $\lim_{(x,y)\to (0,0)}\frac{\sin(xy)}{xy}$ First of all, thanks for any help provided.  My question is how to properly solve this limit:
$\lim_{(x,y)\to (0,0)}\frac{\sin(xy)}{xy}$
I should be 1 as it look, I tried it using polar coordinates and I obtained this limit:
$\lim_{r \to 0} \frac{\sin(r^2\sin(\theta)\cos(\theta))}{r^2\sin(\theta)\cos(\theta)}$ 
where I am using $x=r\cos(\theta)$ and $y=r\sin(\theta)$.  From this limit how I conclude that is equal 1?  I guess $\theta$ don't approach any value while $x,y \to 0$ and that is because I didn't wrote it in the limit (is that correct?).
Thank you

If $z:=xy$ and $(x,y)\to(0,0)$, then $z\to 0$. Hence
$$
\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{xy} = \lim_{z\to0}\frac{\sin(z)}{z}=1.
$$
You can also use your approach with polar coordinates as well. As $r\to0$, despite the fact that $\theta$ depends on $r$, we still have $r^2\sin(\theta)\cos(\theta)\to0$ by the squeeze theorem (because $|\sin(\theta)\cos(\theta)|\leq 1$ and $r^2\to0$). Then taking $z:=r^2\sin(\theta)\cos(\theta)$ as above will yield the same result.
