Answer
(a) Yes, it is possible by defining the function $f\left( {x,y} \right)$ such that
$f\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sin \left( {xy} \right)}}{{xy}}}&{if}&{xy \ne 0}\\
1&{if}&{xy = 0}
\end{array}} \right.$
(b) As can be seen from the graph attached, near the axes, the values of $f\left( {x,y} \right)$ are approaching $1$. It supports our conclusion in part (a).
Work Step by Step
(a) We can define the function $f\left( {x,y} \right)$ such that
$f\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sin \left( {xy} \right)}}{{xy}}}&{if}&{xy \ne 0}\\
1&{if}&{xy = 0}
\end{array}} \right.$
First we prove that $\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\sin \left( {xy} \right)}}{{xy}} = 1$
Write $u = xy$. So, the limit becomes
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\sin \left( {xy} \right)}}{{xy}} = \mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{u}$
Using L'Hôpital's Rule, we obtain
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\sin \left( {xy} \right)}}{{xy}} = \mathop {\lim }\limits_{u \to 0} \frac{{\sin u}}{u} = \mathop {\lim }\limits_{u \to 0} \frac{{\cos u}}{1} = 1$
Since $\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\sin \left( {xy} \right)}}{{xy}} = 1 = f\left( {0,0} \right)$, by definition $f\left( {x,y} \right)$ is continuous function.
(b) As can be seen from the graph attached, near the axes, the values of $f\left( {x,y} \right)$ are approaching $1$. Thus, the result support our conclusion in part (a).
