Proof of the fact that $\ln(a) = f '(0)$ for $f(x) = a^x$? Looking over notes from class today and wanted to know if there is any type of proof for the fact that $\ln(a) = \lim_{h\to0}(a^h-1)/h$, which is just $f '(0)$ for any function of the form $f(x) = a^x$. I see that it simply is the case but where's the "mathy" proof of it? Thanks.
 A: Hint:
Use the definition 
$$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}
\\=\lim_{h\to 0}\frac{a^h-1}{h}\\=\lim_{h\to 0}\frac{e^{\ln a^h}-1}{h}\\=\lim_{h\to 0}\frac{e^{h\ln a}-1}{h}\\=\lim_{h\to 0}\frac{1+h\ln a+\cdots-1}{h}=\ln a$$ 
A: First use the commonly known limit "$\lim \frac{e^h-1}{h}$", that is:
$$1 = \lim_{x \rightarrow 0} \frac{e^x - 1}{x}
\\ = \lim_{\log(a)x \rightarrow 0} \frac{e^{\log(a)\cdot x} - 1}{\log(a)\cdot x}
\\ = \lim_{x \rightarrow 0} 1/\log(a)\cdot \frac{a^x - 1}{x}$$
and so $f'(0) = \log(a)$ by the definition of the derivative.
A: Look at it this way: $$f(x)=a^x \implies f'(x)=a^x \ln a$$ this follows from the fact that of $y=a^x$ then $\ln y =x \ln a$ and so $1/y \cdot y' = \ln a \implies y'=a^x \ln a$. So we proved the derivative in other words (using the definition of $e$ as the exponential function base which has a derivative equal to the function itself). 
There is another way to go about finding this derivative, which is to use the definition of the derivative: $$f(x)=a^x \implies f'(x)= \lim_{h\rightarrow 0} \frac{a^{x+h}-a^x}{h}=\lim_{h \rightarrow 0} \frac{a^x(a^h-1)}{h}=a^x \cdot \lim_{h \rightarrow 0} \frac{a^h-1}{h}$$ Now set the two derivatives equal to one another: $$a^x \ln a = a^x \cdot \lim_{h \rightarrow 0} \frac{a^h-1}{h} \implies \ln a= \lim_{h \rightarrow 0} \frac{a^h-1}{h}$$
