Natural Logarithm expressed as Limit

Question

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.

Answer 1

Hint you can prove it using basic limit theorem( exponential limit) ...

(1+$\frac{1}{x})^x$=e . By assuming $ a^x-1=y$then you may see as x→0,y→0.then log(y+1)/loga=x .then substituting and using the exponential limit you can prove it

Answer 2

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 $C(a) a^x$.