Answer
\[1\]
Work Step by Step
\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} \hfill \\
{\text{Making a table of values and using a calculator}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
x&{\frac{{\ln \left( {1 + x} \right)}}{x}}&x&{\frac{{\ln \left( {1 + x} \right)}}{x}} \\
{{{10}^{ - 1}}}&{0.953102}&{ - {{10}^{ - 1}}}&{1.053605} \\
{{{10}^{ - 2}}}&{0.995033}&{ - {{10}^{ - 2}}}&{1.005034} \\
{{{10}^{ - 3}}}&{0.999500}&{ - {{10}^{ - 3}}}&{1.000500} \\
{{{10}^{ - 4}}}&{0.999950}&{ - {{10}^{ - 4}}}&{1.000050} \\
{{{10}^{ - 5}}}&{0.999995}&{ - {{10}^{ - 5}}}&{1.000005} \\
{{{10}^{ - 6}}}&{1.000000}&{ - {{10}^{ - 6}}}&{1.000001} \\
{{{10}^{ - 7}}}&{1.000000}&{ - {{10}^{ - 7}}}&{1.000000}
\end{array}} \hfill \\
\hfill \\
{\text{Find the limit using the L'Hopital's Rule}} \hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {\ln \left( {1 + x} \right)} \right]}}{{\frac{d}{{dx}}\left[ x \right]}} \hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{1 + x}}}}{{\left( 1 \right)}} \hfill \\
= \mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + x}} \hfill \\
{\text{Evaluating the limit}} \hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + x}} = \frac{1}{{1 + 0}} \hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + x}} = 1 \hfill \\
\end{gathered} \]
