Prove that $\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$

Question

Without using L'Hopital's rule, prove that $$\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$$

Answer 1

Don't know if that is rigorous enough for you, but this is one way using $e^t = \sum_{k=0}^{\infty} \frac{t^k}{k!}$, $e^{\ln x} = x$ and some results about absolute convergent series:

\begin{align*} \frac{a^x - 1}{x} & = (e^{x \cdot \ln a} -1)/x = \left(\sum_{k=0}^{\infty} \frac{(x \cdot \ln a)^k}{k!} - 1 \right)/x\\ & = \left( x \cdot \ln a + \frac{(x \cdot \ln a)^2}{2} + \ldots \right)/x = \ln a + (\ln a)^2 \cdot \frac{x}{2} + (\ln a)^3 \cdot \frac{x^2}{6} +\ldots\\ \end{align*}

Hence, $\lim_{x \to 0} \frac{a^x -1}{x} = \ln a$.