Answer
$$\ln 10$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{{{10}^h} - 1}}{h} \cr
& {\text{The limit can be written as }} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{10}^h} - {{10}^0}}}{{h - 0}} \cr
& {\text{From the definition of the derivative }} \cr
& f'\left( c \right) = \mathop {\lim }\limits_{h \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}} \cr
& \underbrace {\mathop {\lim }\limits_{h \to 0} \frac{{{{10}^h} - {{10}^0}}}{{h - 0}}}_{f'\left( c \right) = \mathop {\lim }\limits_{h \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}}} \Rightarrow f\left( h \right) = {10^h},{\text{ }}a = 0 \cr
& {\text{Then }} \cr
& f'\left( h \right) = \frac{d}{{dh}}\left[ {{{10}^h}} \right] \cr
& f'\left( h \right) = {10^h}\left( {\ln 10} \right) \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{10}^h} - {{10}^0}}}{{h - 0}} = f'\left( 0 \right) \cr
& f'\left( 0 \right) = {10^0}\left( {\ln 10} \right) = \ln 10 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{{{10}^h} - 1}}{h} = \ln 10 \cr} $$
