In the process of trying to prove the derivative of $f(x)=a^x$ (for $a\in\mathbb{R}$) using the definition of the derivative, one arrives at the following equation:
\begin{align} \frac{df}{dx} = \frac{d}{dx}\left[a^x\right] = \lim_{\Delta x \rightarrow 0}\left[\frac{a^{(x+\Delta x)}-a^x}{\Delta x}\right] = \lim_{\Delta x \rightarrow 0}\left[\frac{(a^{\Delta x} – 1)a^x}{\Delta x}\right] \end{align}
At this point, I wish to show that
\begin{align} \lim_{\Delta x \rightarrow 0}\left[\frac{a^{\Delta x} – 1}{\Delta x}\right] = \ln(a). \end{align}
How can one show that this is true in this context WITHOUT Taylor Series and WITHOUT the knowledge of the derivative of $e^x$? (i.e. from first principles in the context of the proof?) L'Hopital's rule seems to be ineffective here since it would involve assuming what we are trying to prove.
Best Answer
In order to show that this limit is $\ln(a)$ you have to bring in the definition of the natural logarithm. And it is not good enough to say that $x = \ln(a) \Leftrightarrow a = e^x$ because that begs the question of how to define $e$.
One typical way to define the natural logarithm is as the integral of $1/x$; but that immediately lets you show that the derivative of $e^x$ is $e^x$ so it may not satisfy your desire.
So I think the best you can do without injecting knowledge of natural logarithms or of $e$, is to prove that the limit you present exists for all positive $a$ (though it might -- does -- depend on $a$) so that the derivative of $a^x$ will be $C(a) a^x$.