Can $\lim_{h\to 0}\frac{b^h - 1}{h}$ be solved without circular reasoning?

Question

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?

Answer 1

There are two standard approaches to the logarithm and exponential.

1. Define the exponential, perhaps via Taylor series or via $\frac{d}{dx} e^x =e^x$. Then, using this, derive properties of its inverse $\ln x$.

2. 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.

Answer 2

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.

Answer 3

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:

$$\lim_{x\to\infty}x\log\left(1+\frac{1}{x}\right)= \lim_{t\to0}\frac{\log(1+t)}{t}= \lim_{t\to0}\frac{\log(1+t)-\log1}{t}= \log'1=1$$