Answer
$$e^2-1$$
Work Step by Step
Given$$\lim _{(x, y) \rightarrow(0,2)} \frac{\left(e^{x}-1\right)\left(e^{y}-1\right)}{x}$$
Since
\begin{align*}
\lim _{(x, y) \rightarrow(0,2)} \frac{\left(e^{x}-1\right)\left(e^{y}-1\right)}{x}&=\lim _{x \rightarrow0} \frac{\left(e^{x}-1\right) }{x}\lim _{y\rightarrow2} \left(e^{y}-1\right) \\
&= (e^2-1) \lim _{x \rightarrow0} \frac{\left(e^{x}-1\right) }{x}\\
& \text{using L'Hopital's Rule } \\
&=( e^2-1) \lim _{x \rightarrow0} \frac{\left(e^{x}\right) }{1}\\
&=e^2-1
\end{align*}
