Answer
\[2\]
Work Step by Step
\[\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{-x}}}{x} \\
& \text{Evaluating the limit directly} \\
& \underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{-x}}}{x}=\frac{{{e}^{0}}-{{e}^{-0}}}{0}=\frac{0}{0} \\
& \text{The limit has the indeterminate form }\frac{0}{0},\text{ then we can apply} \\
& \text{the L }\!\!'\!\!\text{ Hopital }\!\!'\!\!\text{ s Rule} \\
& \text{The Maclaurin series for }{{e}^{x}}\text{ is: }\left( \text{go to page 694}\text{, table 9}\text{.5} \right) \\
& {{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+\cdots +\frac{{{x}^{k}}}{k!}+\cdots \\
& and \\
& {{e}^{x}}=1-x+\frac{{{x}^{2}}}{2!}-\cdots +\frac{{{x}^{k}}}{k!}+\cdots \\
& \text{Substituting into the given limit} \\
& \underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{-x}}}{x} \\
& \underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1+x+\frac{{{x}^{2}}}{2!}+\cdots +\frac{{{x}^{k}}}{k!}+\cdots \right)-\left( 1-x+\frac{{{x}^{2}}}{2!}-\cdots +\frac{{{x}^{k}}}{k!}+\cdots \right)}{x} \\
& \underset{x\to 0}{\mathop{\lim }}\,\frac{1+x+\frac{{{x}^{2}}}{2!}+\cdots +\frac{{{x}^{k}}}{k!}-1+x-\frac{{{x}^{2}}}{2!}-\cdots -\frac{{{x}^{k}}}{k!}+\cdots }{x} \\
& \text{Simplifying} \\
& \underset{x\to 0}{\mathop{\lim }}\,\frac{2x}{x} \\
& \underset{x\to 0}{\mathop{\lim }}\,\left( 2 \right) \\
& \text{Evaluating the limit} \\
& \underset{x\to 0}{\mathop{\lim }}\,\left( 2 \right)=2 \\
\end{align}\]
