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?

  • 2
    $\begingroup$ Possible duplicate of Is $e$ a coincidence? $\endgroup$
    – lioness99a
    Jun 6, 2017 at 13:41
  • 1
    $\begingroup$ Here's a hella long proof of it existing: www2.bc.edu/robert-c-haraway/exist.pdf $\endgroup$
    – Brenton
    Jun 6, 2017 at 13:42
  • $\begingroup$ 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. $\endgroup$
    – Kaynex
    Jun 6, 2017 at 13:46
  • 2
    $\begingroup$ 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 $\lim_{n\to\infty} \left(1+x/n\right)^n=e^x.$ $\endgroup$
    – sharding4
    Jun 6, 2017 at 13:49
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 6, 2017 at 14:56

3 Answers 3


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 $h<0$,


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 $b=e$.

  • $\begingroup$ 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 $a$). $\endgroup$
    – Paramanand Singh
    Jun 8, 2017 at 5:08
  • $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Jun 8, 2017 at 5:11
  • $\begingroup$ @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. $\endgroup$
    – CY Aries
    Jun 8, 2017 at 6:21
  • $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Jun 8, 2017 at 7:03
  • $\begingroup$ @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? $\endgroup$
    – CY Aries
    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.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .