In this YouTube video it is said that $ e $ naturally arises as a number that allows us to take the derivatives of functions like $ a^x $. So $ e $ is defined as a number for which: $$(e^x)'=e^x\lim_{h\rightarrow 0}\frac {e^h-1}{h}=e^x$$
So essentially the limit should be $1 $ for this to happen. How would you prove that there is indeed such a real number, and find its value?
P.S.Please, don't say that there are other ways to define $ e $ - I know that. It's just this particular definition that seems natural and interesting to me.
If you had a definition of $a^b$ (which most books base on having a definition of $\ln$ and $e$, so that $a^b := e^{b \ln a}$), you might say to yourself
$a^b$ depends continuously on $a$ and $b$ as long as $a$ and $b$ are positive.
When I compute $\lim_{h \to 0} \frac{2^h - 1}{h}$, I get a number less than 1.
When I compute $\lim_{h \to 0} \frac{4^h - 1}{h}$, I get a number greater than 1.
Therefore by the intermediate value theorem, there should be a number $c$ with $$ \lim_{h \to 0} \frac{c^h - 1}{h} = 1. $$
This depends, of course, on the well-definedness and continuity of the function
$$ F(b) =\lim_{h \to 0} \frac{b^h - 1}{h} $$
which I cannot prove without either a very messy epsilon-delta argument, or relying on alternative definitions of $e$. But this is at least the gist of how you might go about things. I'd love to see your definition of $a^b$ for irrational $a$ and $b$, though, so that I could turn it into a real proof.