In many places I have read that $$\lim_{h\to 0}\frac{b^h - 1}{h}$$ is by definition $\ln(b)$. Does that mean that this is unsolvable without using that fact or a related/derived one?
I can of course solve it with L'Hospitals rule, but that uses the derivative of the exponential function which is exactly what I want to derive by solving this limit.
Since the derivative of the logarithm can be derived from the derivative of the exponential, using the fact that they are inverses, means that deferring this limit to something that can be solved using the derivative of log seams also cheating.
Others have asked that before, but all the "non-L'Hospital"-solutions seem to defer it to some other limit that they claim obvious. For example two solutions from Proving that $\lim_{h\to 0 } \frac{b^{h}-1}{h} = \ln{b}$ use $$\lim_{x\to 0}\frac{\log_a(1+x)}{x}=\log_a e$$ and $$\lim_{x\to 0}\frac{e^x-1}x=1$$
none of which is more obvious (to me) as the original.
On that same page, the 1st of the above 2 is "derived" using a Taylor expansion (if I am not mistaken) which (if I remember correctly) is based on derivatives (in this case of the logarithm), which is related to the derivative of exp as I meantioned above. So this seems to be circular reasoning too. (a very large circle though)
So is this limit not solvable at all without using smth that is based on something this meant to prove? Can this only be defined to equal $\ln(b)$; and numerically determined to some precision?
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