How does one prove that $e$ exists?

Solution 1:

Let $a>0$ and $a\ne1$. First we have to prove the existence of $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$.

Assume that $r>1$ and let $f(x)=x^r-rx+r-1$ for $x>0$. Then

$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }0<x<1\\ =0 &\text{if }x=1\\ >0 &\text{if }x>1 \end{cases}$$

Therefore, $f$ attains its absolute minimum at $x=1$. So for all $x>0$, we have

$$f(x)\ge f(1)=0$$

$$x^r\ge rx+1-r$$

So when $r>1$ and $h>0$, $\displaystyle\frac{a^{rh}-1}{rh}\ge\frac{ra^h+1-r-1}{rh}$ and hence

\begin{align} \frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\ge0 \end{align}

When $r>1$ and $h<0$, $\displaystyle\frac{a^{rh}-1}{rh}\le\frac{ra^h+1-r-1}{rh}$ and hence

\begin{align} \frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\le0 \end{align}

Therefore, $\displaystyle \frac{a^h-1}{h}$ is an increasing function in $h$. As it is bounded below by $0$ on $(0,\infty)$, $\displaystyle \lim_{h\to0^+}\frac{a^h-1}{h}$ exists.

When $h<0$,

$$\frac{a^h-1}{h}=a^h\left(\frac{a^{-h}-1}{-h}\right)$$

As $\displaystyle \lim_{h\to0^-}a^h$ exists and equals $1$, $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}$ exists and $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}= \lim_{h\to0^+}\frac{a^h-1}{h}$.

Therefore, $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$ exists.

Now we are ready to prove that there exists an $e$ such that $\displaystyle \lim_{h\to0}\frac{e^h-1}{h}=1$.

Define $e=a^\frac{1}{k}$, where $\displaystyle k=\lim_{h\to0}\frac{a^h-1}{h}$. Then

\begin{align} \lim_{h\to0}\frac{e^h-1}{h}&=\lim_{h\to0}\left(\frac{a^\frac{h}{k}-1}{\frac{h}{k}}\cdot \frac{1}{k}\right)\\ &=k\cdot\frac{1}{k}\\ &=1 \end{align}

This number $e$ is unique. Indeed, if $b>0$ and $\displaystyle \lim_{h\to0}\frac{b^h-1}{h}=1$, then we can prove that $b=e$.

Let $p=\log_eb$. Then $b=e^p$.

\begin{align} \lim_{h\to 0}\frac{b^h-1}{h}-\lim_{h\to 0}\frac{e^h-1}{h}&=1-1\\ \lim_{h\to 0}\frac{e^{ph}-e^h}{h}&=0\\ \lim_{h\to 0}\left[(p-1)e^h\cdot\frac{e^{(p-1)h}-1}{(p-1)h}\right]&=0\\ (p-1)(1)(1)&=0\\ p&=1 \end{align}

Hence $b=e$.

Solution 2:

It depends on how you define $e$. In some senses, you could define $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$ and then from here set $x=1$ and you can prove that $e$ has all the properties we know and love. You can also show that this is equivalent to the limit definition of $e$.

It exists because its definition is based on concepts that are already well-defined (and the power series is convergent for all $x \in \mathbb{R}$ and so defines a function on $\mathbb{R}$).