This limit is actually the definition of the derivative of $a^x$ at $x=0$. Since $\frac{d}{dx}(a^x) = a^x \ln a$, at $x=0$ it becomes $a^0 \ln a = \ln a$.
2. Apply L'Hôpital's Rule: Differentiate the numerator and the denominator with respect to $x$:
$$\frac{d}{dx}(a^x - 1) = a^x \log a$$
$$\frac{d}{dx}(x) = 1\lt strong\gt 3. Evaluate the Limit:\lt /strong\gt \lim_{x \to 0} \frac{a^x \log a}{1} = a^0 \log a$$
Since $a^0 = 1$:
$$\text{Limit} = 1 \cdot \log a = \log a$$
The base of the logarithm is $e$ (natural logarithm), often written as $\ln a$.