Answer
The limit
$f\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{2^x} - 1} \right)\left( {\sin y} \right)}}{{xy}}}&{if}&{xy \ne 0}\\
{\ln 2}&{if}&{xy = 0}
\end{array}} \right.$
is continuous at $\left( {0,0} \right)$.
Work Step by Step
Firstly, we find the limit
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left( {{2^x} - 1} \right)\left( {\sin y} \right)}}{{xy}}$
Since the limit is equal to a product of limits, so
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left( {{2^x} - 1} \right)\left( {\sin y} \right)}}{{xy}} = \left( {\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}} \right)\left( {\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y}} \right)$
But $\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y} = 1$, so
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left( {{2^x} - 1} \right)\left( {\sin y} \right)}}{{xy}} = \left( {\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}} \right)\left( {\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}$
Recall from Theorem 1 of Section 7.3 on page 338, we obtain the derivative of ${2^x}$ with respect to $x$:
$\frac{d}{{dx}}{2^x} = \left( {\ln 2} \right){2^x}$
Applying L'Hôpital's rule to the limit, we get
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left( {{2^x} - 1} \right)\left( {\sin y} \right)}}{{xy}} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\ln 2} \right){2^x}}}{1} = \ln 2$
Thus,
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left( {{2^x} - 1} \right)\left( {\sin y} \right)}}{{xy}} = \ln 2 = f\left( {0,0} \right)$
Hence, by definition the function is continuous at $\left( {0,0} \right)$.
