TITLE: Proving uniqueness of $e$
QUESTION [5 upvotes]: Let's define $e$ as the number $a$ such that $\frac {d}{dx} a^x = a^x$.  I'm trying to prove that this $a$ has to be $e$.  I don't see any way of proceeding from here except by the limit definition (I'm not assuming I know what the $\ln$ function is, or else there'd be a much easier definition of $e$ to be had).
$$\frac {d}{dx} a^x=\lim_{h\to 0} \frac {a^{x+h}-a^x}{h}=\lim_{h\to 0} a^x\frac{a^h-1}{h}=a^x\left(\lim_{h\to 0} \frac{a^h-1}{h}\right)$$
So clearly $e$ must be the number that makes that limit on the far right equal to $1$.  I'm not sure how to evaluate this. Using L'Hopital's rule, I just get $\lim_{h\to 0} \frac {d}{dh} a^h$, which doesn't particularly help.
So my question:

How can I prove that there is a unique number such that $\lim_{h\to 0} \frac {a^h-1}{h}=1$?

REPLY [3 votes]: First, note that $\frac {d}{dx} 0^x = 0^x$, so you really want to know that there is a unique positive real number $a$ for which $\frac {d}{dx} a^x = a^x$.
Suppose that $\frac{d}{dx}a^x=a^x$ and $\frac{d}{dx}b^x=b^x$, with $a$ and $b$ both positive. Then by the quotient rule,
$$\frac{d}{dx}\left(\frac{a}{b}\right)^x=\frac{d}{dx}\frac{a^x}{b^x}=\frac{b^xa^x-a^xb^x}{b^{2x}}=0.$$ Then $\left(\frac{a}{b}\right)^x$ is a constant, which implies that $\frac{a}{b}$ is $1$ or $0$. Since $a>0$, $\frac{a}{b}$ must be $1$, so $a=b$.