Natural Logarithm expressed as Limit 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.

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
