Answer
$\ln a $
Work Step by Step
We have
$$
\lim _{x \rightarrow 0}\frac{a^x-1}{x}=\frac{0}{0}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0}\frac{a^x-1}{x}=\lim _{x \rightarrow 0}\frac{a^x \ln a}{1}=\ln a.
$$
