Continuity of a function at $(0,0)$ Is $f$ continuous at $(0,0)$?
$$f(x, y) := \begin{cases}\frac{\sin(xy)}{|x|+|y|},&\ \text{ if }(x, y)\ne(0,0\\ \ \\ 0,&\ (x,y)=(0,0)\end{cases}
$$
My attempt:
We know $|xy| \leq \frac{1}{2}(x^2 + y^2)$ and so $-\frac{1}{2}(x^2 + y^2) \leq xy\leq \frac{1}{2}(x^2 + y^2)$.
Thus $$- \sin(\frac{1}{2}(x^2 + y^2))\leq \sin(xy) \leq  \sin(\frac{1}{2}(x^2 + y^2)).$$
And hence, $$\frac{\sin(\frac{1}{2}(x^2 + y^2)}{|x| + |y|}\leq\frac{\sin(xy)}{|x| + |y|}\leq-\frac{\sin(\frac{1}{2}(x^2 + y^2)}{|x| + |y|}.$$
Now using polar coordinates, it becomes
$$\frac{\sin(\frac{1}{2}r^2)}{r(\cos \theta + \sin \theta)} \leq\cdots$$
So by sandwich theorem, 
$\frac{\sin(\frac{1}{2}r^2)}{r(\cos\theta + \sin\theta)}$ tends to $0$ as $r$ tends to $0$, so $f$ is continuous.
Is my attempt correct?
Any other suggestions please? 
 A: It's hard for me to see what you are trying to achieve. The sine is not monotone everywhere, so you need to be careful if you apply it. You seem to reverse an inequality there, and I don't see why. 
And, if you are going to use polar coordinates, you can use them directly, you gain nothing by the inequalities you try to use first. 
Finally,  you don't really need polar coordinates: using the easy inequality $|\sin t|\leq|t|$, 
\begin{align}
\left|\frac{\sin(xy)}{|x|+|y|}\right|&=\frac{|\sin xy|}{|x|+|y|}\leq\frac{|xy|}{|x|+|y|}
\leq\frac12\,\frac{x^2+y^2}{|x|+|y|}\\ \ \\
&=\frac12\,\left(\frac{x^2}{|x|+|y|}+\frac{y^2}{|x|+|y|} \right)\\ \ \\
&\leq \frac12\,\left( \frac{x^2}{|x|}+\frac{y^2}{|y|}\right)\\ \ \\
&=\frac12\,(|x|+|y|).
\end{align}
The estimate can actually be made a lot simpler (maybe switching the roles of $y$ and $x$ if $x=0$): 
\begin{align}
\left|\frac{\sin(xy)}{|x|+|y|}\right|&=\frac{|\sin xy|}{|x|+|y|}\leq\frac{|xy|}{|x|+|y|}=\frac{|x|\,|y|}{|x|+|y|}\leq\frac{|x|\,|y|}{|x|}=|y|.
\end{align}
Note that if $x=0$ or $y=0$, then no inequalities are needed as already $\sin xy=0$. 
