How to prove that $\lim\limits_{h \to 0} \frac{a^h - 1}{h} = \ln a$ In order to find the derivative of a exponential function, on its general form $a^x$ by the definition, I used limits.
$\begin{align*}
\frac{d}{dx} a^x & = \lim_{h \to 0} \left [ \frac{a^{x+h}-a^x}{h} \right ]\\  \\
& =\lim_{h \to 0} \left [ \frac{a^x \cdot a^h-a^x}{h} \right ]
\\  \\ 
&=\lim_{h \to 0} \left [ \frac{a^x \cdot (a^h-1)}{h} \right ]
\\  \\
&=a^x \cdot \lim_{h \to 0} \left [\frac {a^h-1}{h} \right ]
\end{align*}$
I know that this last limit is equal to $\ln(a)$ but how can I prove it by using basic Algebra and Exponential and Logarithms properties?
Thanks

It depends a bit on what you're prepared to accept as "basic algebra and exponential and logarithms properties". Look first at the case where $a$ is $e$. You need to know that $\lim_{h\to0}(e^h-1)/h)=1$. Are you willing to accept that as a "basic property"? If so, then $a^h=e^{h\log a}$ so $$(a^h-1)/h=(e^{h\log a}-1)/h={e^{h\log a}-1\over h\log a}\log a$$ so $$\lim_{h\to0}(a^h-1)/h=(\log a)\lim_{h\to0}{e^{h\log a}-1\over h\log a}=\log a$$
