Answer
\[0.693147\]
Work Step by Step
\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} \hfill \\
{\text{Making a table of values and using a calculator}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
x&{\frac{{{2^x} - 1}}{x}}&x&{\frac{{{2^x} - 1}}{x}} \\
{{{10}^{ - 1}}}&{0.717735}&{ - {{10}^{ - 1}}}&{0.669670} \\
{{{10}^{ - 2}}}&{0.695555}&{ - {{10}^{ - 2}}}&{0.690750} \\
{{{10}^{ - 3}}}&{0.693387}&{ - {{10}^{ - 3}}}&{0.692907} \\
{{{10}^{ - 4}}}&{0.693171}&{ - {{10}^{ - 4}}}&{0.693123} \\
{{{10}^{ - 5}}}&{0.693150}&{ - {{10}^{ - 5}}}&{0.693145} \\
{{{10}^{ - 6}}}&{0.693147}&{ - {{10}^{ - 6}}}&{0.693147} \\
{{{10}^{ - 7}}}&{0.693147}&{ - {{10}^{ - 7}}}&{0.693147}
\end{array}} \hfill \\
\hfill \\
{\text{Find the limit using the L'Hopital's Rule}} \hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{2^x} - 1} \right]}}{{\frac{d}{{dx}}\left[ x \right]}} \hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\left( {\ln 2} \right)}}{{\left( 1 \right)}} \hfill \\
{\text{Evaluating the limit}} \hfill \\
\mathop {\lim }\limits_{x \to 0} {2^x}\left( {\ln 2} \right) = \left( {\ln 2} \right)\left( {{2^0}} \right) \hfill \\
\mathop {\lim }\limits_{x \to 0} {2^x}\left( {\ln 2} \right) = \ln 2 \approx 0.693147 \hfill \\
\end{gathered} \]
