# How does one prove that $e$ exists?

In my calculus class, $e$ was defined to be the number such that $\frac{d}{dx}e^x = e^x$.

From the definition of the derivative, we have

\begin{align*} \frac{d}{dx}a^x &= \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\\ &= a^x \lim_{h \to 0} \frac{a^h - 1}{h} \end{align*}

Thus $e$ is the number such that

$$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$$

But how is it proven that there exists such number?

• Possible duplicate of Is $e$ a coincidence? Jun 6, 2017 at 13:41
• Here's a hella long proof of it existing: www2.bc.edu/robert-c-haraway/exist.pdf Jun 6, 2017 at 13:42
• Rather then treating it as a limit, pretend $h$ is "very small" and solve the equation for $e$. It's not rigorous, but this should give you the common definition of $e$. Now, all you would have to do is prove it is a limit that exists. Jun 6, 2017 at 13:46
• You can show that $\lim_{n\to\infty} \left(1+1/n\right)^n$ exists and call this limit $e$. Then $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ follows from that and $\lim_{n\to\infty} \left(1+x/n\right)^n=e^x.$ Jun 6, 2017 at 13:49
• How do you define $a^h$ and $a^x$ if you don't have $e^x$? Usually you start with $e^x$, define $\log x$ from that (or vice versa), then define $a^x=e^{x \log a}$. There are a number of ways to define $e^x$. You pick one, then prove the rest as theorems. If you are going to follow your approach, you have to prove the function $e^x$ exists and is unique from the definition, then $e=e^1$ is easy. You need $e^0=1$ as part of your definition. Jun 6, 2017 at 14:56

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

$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }00 &\text{if }x>1 \end{cases}$$

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 $$b=e$$.

• Nice proof about existence of $e$ so +1 but you need to show that the $e$ defined in this manner is unique (ie independent of the number $a$). Jun 8, 2017 at 5:08
• Also your starting line does it seem to be related to rest of your answer. Your answer does not in anyway depend on log function. Jun 8, 2017 at 5:11
• @ParamanandSingh I originally planned to define $e$ by logarithm. That's why I have put the first line. It is deleted now. Thanks for pointing out the uniqueness problem. I have added the proof in my answer. Jun 8, 2017 at 6:21
• The part about uniqueness has some issue. Since $e$ is not unique you are not allowed to take logs to the base $e$. Uniqueness follows by the following result $$\lim_{h\to 0}\frac{a^{h}-1}{h}=0\Rightarrow a=1$$ you should try to prove this. One way is to show that the limit defines a function of $a$ which is strictly increasing. Jun 8, 2017 at 7:03
• @ParamanandSingh I starts with an arbitrary $a$ and define $e$ by $e=a^{1/k}$. This $e$ is not guaranteed to be unique. But then I suppose that there is a $b$ such that $\lim_{h\to0}\frac{b^h-1}{h}=1$ and let $p$ be the logarithm of this $b$ to this particular $e$ (I am not assuming the uniqueness of $e$ here). And I finally proved that $p=1$ and conclude that $b=e$. So I essentially proved that when $b$ has the property that $\lim_{h\to0}\frac{b^h-1}{h}=1$, this $b$ must be equal $e$. Is this approach correct? Jun 8, 2017 at 7:50

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 $\mathbb{R}$).

Here's a proof for the alternative form $$e := \lim_n (1+1/n)^n$$.

Note first that the binomial theorem yields

$$e = \lim_n \sum_{k=0}^n \binom{n}{k} \left ( \frac{1}{n} \right )^k$$

and the elementary bounds $$\frac{n^k}{k^k} \le \binom{n}{k} \le \frac{n^k}{k!}$$ yield respective bounds

$$\lim_n \sum_{k=0}^n \frac{1}{k^k} \le e \le \lim_n \sum_{k=0}^n \frac{1}{k!}$$

Fixing any $$n > 1$$ and avoiding the limit for the lower bound gives the weaker bound $$2 < e$$, while noting that $$k! \ge 2^{k-1}$$ for $$k \ge 1$$ gives the weaker bound $$e \le 1 + \lim_n \sum_{k=0}^n 2^{-k} = 3$$, i.e., $$2 < e \le 3$$.

(Note that the sophomore's dream has made an appearance here.)