When applying the definition of a derivative to $\frac{d}{dx}b^x$ and a little algebra one arrives to
$b^x\times\lim\limits_{h \to 0}\frac{b^h - 1}{h}$
where of course that limit equals $\ln(b)$. I know this limit can be evaluated with L'Hospitals rule, but that involves using the derivative what I am just about to prove, so that would be a circular proof. I suspect one has to reduce this limit to something that relates to the definition of e as $\lim\limits_{n \to \infty}(1+1/n)^n$ but I can not do it.
How to show that that limit equals $\ln(b)$, so the proof of the derivative of $b^x$ is complete ?
Note: I found that by substituting for $s = b^h-1$, etc, one can separate $\ln(b)$, but than the another indeterminate limit remains: $\lim\limits_{s \to 0}(s/\ln(1+s))$, which must be $1$, and again I do not want to solve it using L'Hospital's rule, but I cannot otherwise.