<s>[INST] Prove that $\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$ Without using L'Hopital's rule, prove that 
$$\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$$
 [/INST] 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, $\lim_{x \to 0} \frac{a^x -1}{x} = \ln a$.
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