Does this definition of $e$ even make sense?

Question

This sprung from a conversation here. In Stewart's Calculus textbook, he defined $e$ as the unique solution to $\lim\limits_{h\to 0}\frac{x^h-1}{h}=1$. Ahmed asked how do you define $x^h$ is not by $\exp(h\ln(x))$ and I'm not sure.

Does this definition of $e$ even make sense?

Definition here:

Answer 1

The definition of $e$ as the unique number such that $$\lim_{h \to 0}\frac{e^{h} - 1}{h} = 1$$ makes sense, but there are few points which must be established before this definition can be used:

1. Define the general power $a^{x}$ for all $a > 0$ and all real $x$. One approach is to define it as the limit of $a^{x_{n}}$ where $x_{n}$ is a sequence of rational numbers tending to $x$ (this is not so easy).
2. Based on the definition of $a^{x}$ above show that the limit $(a^{x} - 1)/x$ as $x \to 0$ exists for all $a > 0$ (this is hard) and hence the limit defines a functions $f(a)$ for $a > 0$.
3. The function $f(x)$ defined above is continuous, strictly increasing and maps $(0, \infty)$ to $(-\infty, \infty)$ (easy if previous points are established).

From the last point above it follows that there is a unique number $e > 1$ such that $f(e) = 1$. This is the definition of $e$ with which we started. And as can be seen this definition must be preceded by the proof of the results mentioned in three points above. All this is done in my blog post and in my opinion this is the most difficult route to a theory of exponential and logarithmic functions. Easier routes to the theory of exponential and logarithmic functions are covered in this post and next.

Answer 2

$b^x$ can be defined as $\lim_{q\in \mathbb Q \rightarrow x}b^q$. (Isn't it usually so defined?) Or alternatively one can define $e = \lim_{h\in \mathbb Q\rightarrow 0}\frac {x^h - 1}{h}$.

I think it's legit and not circular.