How can we prove that there is a number $a$ such that $\lim_{h\to 0}\frac{a^{h} -1}{h}=1$? One definition of $e$ that I am fond of is that it is the number $a$ such that
$$
\lim_{h\to 0}\frac{a^{h} -1}{h}=1
$$
The reason for this is that it cuts to the heart of the special property of all exponential functions. If we have $f(x)=a^x$, then
\begin{align}
f'(x)&=\lim_{\Delta x \to 0}\frac{a^{x+\Delta x} -a^x}{\Delta x} \\
&=\lim_{\Delta x \to 0}a^x\frac{a^{\Delta x} - 1}{\Delta x} \\
&=a^x\lim_{\Delta x \to 0}\frac{a^{\Delta x} - 1}{\Delta x}
\end{align}
However, presumably there is a caveat to this approach. We need to show that a number like $e$ exists in the first place! In other words, we need to show that $g(a)=\lim_{\Delta x \to 0}\frac{a^{\Delta x} -1}{\Delta x}$ takes the value of $1$ for some value of $a$. How might we do this?
 A: Independently, we can prove that there exists a number $e$ such that $e = \lim_{n \to \infty} (1+1/n)^n$ along with the property that for all $n \in \mathbb{N}$,
$$\left(1 + \frac{1}{n} \right)^n < e < \left(1 + \frac{1}{n} \right)^{n+1}$$
It follows that
$$1 < n(e^{1/n}-1) < n\left[\left(1 + \frac{1}{n} \right)^{1/n}\left( 1+ \frac{1}{n}\right)-1  \right] \leqslant 1 + \frac{1}{n} + \frac{1}{n^2},$$
where the right-hand inequality is obtained using Bernoulli's inequality $(1 + 1/n)^{1/n} \leqslant 1 + 1/n^2$.
By the squeeze theorem we get
$$\tag{*}\lim_{n \to \infty}\frac{e^{1/n}-1}{\frac{1}{n}} = 1$$
From here it is not difficult to show that
$$\tag{**} \lim_{h \to 0+}\frac{e^{h}-1}{h} = 1$$
Taking $n = \lfloor1/h\rfloor$ when $h > 0$, we have $n \leqslant 1/h < n+1$ and
$$\frac{n}{n+1}(n+1)(e^{1/n+1} - 1)=  n(e^{1/n+1} - 1 ) \leqslant \frac{e^h -1 }{h} \leqslant (n+1)(e^{1/n} -1) = \frac{n+1}{n}n(e^{1/n}-1)$$
Since $n \to \infty$ if and only if  $h \to 0$ we obtain (**) by the squeeze theorem using the previous result (*).
With some more work we can show that the limit $1$ is also attained as $h \to 0-$.
A: Let $f(x)$ be given by the
$$f(x)=\int_1^x \frac1t\,dt$$
It is easy to show that $f(x)$ is continuous and increasing for $x>0$ with $f(1)=0$ and $\lim_{x\to\infty}f(x)=\infty$.  Then, by the intermediate value theorem, there exists a number $a>1$ such that
$$f(a)=\int_1^a \frac1t\,dt=1\tag1$$

We enforce the substitution $t\mapsto (1+ht)^{1/h}$ in $(1)$ to obtain
$$1=\int_0^{(a^h-1)/h}\frac1{1+ht}\,dt\tag2$$

For $h>0$, $\frac{a^h-1}{h}>0$ and $1\le 1+ht\le a^h$ when $t\in [0,(a^h-1)/h]$.  Therefore, from $(2)$ we find that
$$\left(\frac{a^h-1}{h}\right)a^{-h}\le 1\le \left(\frac{a^h-1}{h}\right)\tag3$$
Rearranging $(3)$, we have the bounds
$$1\le \left(\frac{a^h-1}{h}\right)\le a^h$$
whence application of the squeeze theorem yields
$$\lim_{h\to 0^+}\left(\frac{a^h-1}{h}\right)=1$$
We leave it as an exercise for the reader to show that the left-side limit is also $1$ from which we conclude
$$\lim_{h\to 0}\left(\frac{a^h-1}{h}\right)=1$$
for some $a>1$ such that $1=\int_1^a \frac1t\,dt$.
A: Let me suggest one more, if it can be so called, functional, approach. Let's consider function $f$ with following 3 property:

*

*$f(x_1+x_2)=f(x_1)f(x_2)$ for $\forall x_1,x_2 \in \mathbb{R}$

*$f(0)=1,f(1)=a$ where $a>0$

*$f$ is continuous in $x=0$
It can be proved, that exists unique continuous on all $\mathbb{R}$ function, which satisfies brought properties and it we define as $a^x$.
Can be proved, that exists limit
$$\lim\limits_{x \to \infty}\left(  1+\frac{1}{x} \right)^x= \lim\limits_{x \to 0}\left(  1+ x\right)^{\frac{1}{x}}$$
and we denote it by $e$. For $e$, at last, we obtain
$$\lim\limits_{x \to 0}\frac{e^x-1}{x}=1$$
If this approach is satisfactory and you are interested in the details, then write which part you would like to see more fully written.
