Does the following limit exist $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y}?$ Does the following limit exist?

$\lim\limits_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y}$

The answer here is not correct in my opinion since it does not take under consideration all the surface parts in which $y=0.$
 A: Set $g(x)=\begin{cases} \frac{x\cdot \sin(xy)}{xy} \quad \text{if}\quad x\neq 0\\ 0 \quad \quad \quad \;\; \text{if}\quad x=0 \end{cases}$
Then $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y}=\lim_{(x,y)\rightarrow (0,0)}g(x)=0,$$ as we can make $\frac{\sin(xy)}{xy}$ as close to $1$ as we wish. 
A: There is a wonderful (entire) function 
$${\rm sinc}(t):=\left\{\eqalign{{\sin t\over t}\quad&(t\ne0)\cr 1\quad\ \ &(t=0)\cr}\right.$$
with ${\rm sinc}(0)=1$, having Taylor series
$${\rm sinc}(t)=1-{1\over3!}t^2+{1\over5!}t^4-\ldots\ ,$$
and satisfying the identity $$t\>{\rm sinc}(t)=\sin t\qquad(t\in{\mathbb R})\ .$$
In terms of this function we have
$${\sin (xy)\over y}={xy\>{\rm sinc}(xy)\over y}=x\>{\rm sinc}(xy)$$
for all points $(x,y)$ where the LHS is defined, i.e., for all $(x,y)$ with $y\ne0$. For this domain we can therefore say that
$$\lim_{(x,y)\to(0,0)}{\sin (xy)\over y}=\lim_{(x,y)\to(0,0)}\bigl(x\>{\rm sinc}(xy)\bigr)=0\cdot1=0\ .$$
A: $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y} = \lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{xy}x =(\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{xy})( \lim _{x\to 0} x)=1\times 0 =0$$
A: Thanks for all the answers!
I came up with another proof:
Proof: Since there exists a $\delta$ neighborhood of $(0,0)$, $\mathcal{B_{\delta}(0,0)}$ , in which $|sin(xy)|<|xy|$ for every $(x,y) \in \mathcal{B_{\delta}(0,0)}$, we can write for every such $(x,y)$:


$|\frac{sin(xy)}{y}|=\frac{|sin(xy)|}{|y|}<\frac{|xy|}{|y|}=|x| \rightarrow 0$ 


What you think, is it ok?
Thanks!
