Answer
$$\ln 10$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{{10}^x} - 1}}{x} \cr
& {\text{evaluating the limit, we get:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{{10}^x} - 1}}{x} = \frac{{{{10}^0} - 1}}{0} \cr
& = \frac{0}{0} \cr
& {\text{the limit is }}\frac{0}{0}{\text{, so}}{\text{ we apply l'Hopital's rule}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( {{{10}^x} - 1} \right)}}{{d/dx\left( x \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{10}^x}\ln 10}}{1} \cr
& = \mathop {\lim }\limits_{x \to 0} \left( {{{10}^x}\ln 10} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& = {10^0}\ln 10 \cr
& = \ln 10 \cr
& {\text{then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{{10}^x} - 1}}{x} = \ln 10 \cr} $$
