Can $\lim_{h\to 0}\frac{b^h - 1}{h}$ be solved without circular reasoning?
Solution 1:
This depends on how you introduce the functions. If you start by defining
$$\log x=\int_1^x \frac{1}{t}\,dt$$
(for $x>0$) then you know from the fundamental theorem that the derivative is $\log' x=1/x$. Thus the function is increasing; it's easy to show the main property $\log(ab)=\log a+\log b$:
$$\int_1^{ab}\frac{1}{t}\,dt= \int_1^a\frac{1}{t}\,dt + \int_a^{ab}\frac{1}{t}\,dt $$
and the substitution $t=au$ in the second integral will give the result. Since $\log2>0$, we get $\log2^n=n\log2$ and this implies that $\lim_{x\to\infty}\log x=\infty$. Since $\log(1/x)=-\log x$ because $\log1=0$, we also get $\lim_{x\to0}\log x=-\infty$. Thus $\log$ takes on all real values and its inverse function $x\mapsto\exp x$ is defined and differentiable on the whole real line, with $\exp' x=\exp x$.
This allows to define $b^x = \exp(b\log x)$ (for $b>0$) and it's easy to show that, for all $x$ and $y$, $b^{(x+y)}=b^xb^y$, which means in particular that
$$b^n=\underbrace{b\cdot b\cdot\dots\cdot b}_{\text{$n$ times}}$$
(easy induction) so the notation $b^x$ is not ambiguous. It's also clear that the derivative of $g(x)=b^x$ is $g'(x)=b^x\log b$, so your limit
$$\lim_{h\to0}\frac{b^h-1}{h}=g'(0)=b^0\log b=\log b$$
without any circular reasoning.
The Napier-Euler number $e$ can be defined by $\log e=1$ (there is a unique solution of this equation because $\log$ is increasing) and by definition we get $\exp x=e^x$. Also the fundamental limit
$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$$
can be computed in the same vein; just take logarithms:
$$\lim_{x\to\infty}x\log\left(1+\frac{1}{x}\right)= \lim_{t\to0}\frac{\log(1+t)}{t}= \lim_{t\to0}\frac{\log(1+t)-\log1}{t}= \log'1=1$$
Solution 2:
There are two standard approaches to the logarithm and exponential.
Define the exponential, perhaps via Taylor series or via $\frac{d}{dx} e^x =e^x$. Then, using this, derive properties of its inverse $\ln x$.
Define the logarithm, perhaps via Taylor series or via $\frac{d}{dx} \ln x = \frac 1x$ or via the limit in the OP. Then, using this, derive properties of its inverse $e^x$.
Similarly, there are multiple approaches to generalizing the logarithm and exponential to different bases, e.g. $b$.
You seem to object to each approach because of the existence of the other approach. The ability to prove both $A\rightarrow B$ and $B\rightarrow A$ does not make either proof circular. As @Andre wrote, you need to choose a definition to get started, and then things won't be circular.
Solution 3:
If we accept
$$\lim_{x\to 0}(1+x)^{1/x}=\text{e}$$
then
$$\lim_{x\to 0}\frac{\log_a(1+x)}{x}=\lim_{x\to 0}\frac{1}{x}\frac{\log_a(1+x)}{x}=\lim_{x\to 0}[\log_a(1+x)]^{1/x}=\lim_{x\to 0}\log_a(1+x)^{1/x}=\log_a\mathrm{e}$$
by applying some known theorems.