# Proving that a definition of e is unique

We can define $e$ as the number such that $\lim_{h \to 0} \frac{e^h-1}{h}=1$. However, of course we can only define $e$ this way if it is unique, i.e., there is no other value $c$ for which that is a true statement. Could someone prove this uniqueness for me? I am tutoring a calculus student and he asked me something to this effect and I couldn't see why. Thanks!

• Offhand, I would consider fiddling with $$\frac{a^h - b^h}{h}$$ to see if the limit can be zero if $a\neq b$. I suspect it cannot. Then just note that $a^h-b^h = (a^h-1)-(b^h-1)$ and go from there.
– MPW
Sep 9, 2014 at 21:32
• Using wiki (analysis is not my domain for sure) for what it is and nothing more, they give a definition that for any other value of $c^x$ (at least this is my interpretation of your question) the gradient of the function is not tangent to $y=x+1$ at $x=0$. But look at this wiki explanation Sep 9, 2014 at 22:43
• Try to show that $(x^h-1) / h$ is increasing with $x$ (can be done with elementary means - no calculus needed). Sep 10, 2014 at 4:03
• @Winther after showing its increasing how do we know that the limit as x->0 exits? Jun 21, 2020 at 13:23

If $x = e^c$ with $c \neq 1$, then $\lim_{h \to 0}\dfrac{x^h - 1}{h} = \lim_{h\to 0}\dfrac{e^{ch} - 1}{ch}c = c \neq 1$

• Is that not circular? Can we use $e^c$ to prove things about the definition of $e$? Sep 9, 2014 at 21:35
• @Mathmo123 we know $e$ satisfies the limit, and we want to prove there is no other number than $e$ satisfying it. I didn't use the uniqueness in the proof Sep 9, 2014 at 21:38
• OK. Also - can you explain how you get $\lim_{h\to 0}\dfrac{e^{ch - 1}}{ch}c$ Sep 9, 2014 at 21:40
• @Mathmo123 It's also because we know $e$ satisfies the limit. $h \to 0$ then $ch\to 0$ Sep 9, 2014 at 21:42
• Should it not be $\lim_{h\to 0}\dfrac{e^{ch}- 1}{ch}c$? Sep 9, 2014 at 21:42

Assume that for some particular number $c > 0$, $c \ne 1$, you know that the limit

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

exists and is finite and nonzero. The fact that a number $c$ with this property exists is by no means obvious, and a proof is, I believe, beyond the reach of a beginning calculus student in 99.9% of cases. (If you really want a proof, it is possible to give one that is very elementary but long, another that uses continuity and properties of convex functions, and another, much shorter one that starts off the entire theory by defining $\ln x$ as the integral of $1/x$.) Pedagogically, it might be preferable to simply assume the existence of $l_c$, say, for $c = 2$.

Nonetheless, if we accept this fact, the existence and uniqueness of $e$ can be established by the following argument.

Let $d = c^r$ for some nonzero real number $r$. (Any positive number $d \ne 1$ can be written in this form.) Then the limit

$l_d = \lim_{h \to 0} \frac{d^h - 1}{h} = \lim_{h \to 0} r\frac{c^{rh} - 1}{rh}$

exists and can be evaluated by making the substitution $k = rh$. Since $k \to 0$ as $h \to 0$, we have

$l_d = \lim_{k \to 0} r\frac{c^{k} - 1}{k} = rl_c$.

We have $l_d = 1$ if and only if $r = 1/l_c$. Thus the only number with the desired property is $e = c^{1/l_c}$.

Note that the actual value of $l_c$ later turns out to be $\ln c$.

I have presented the theory of exponential and logarithmic functions in my blog post. For the current question we neeed to understand that for each $a>0$, the limit $$\lim_{h\to 0}\frac{a^{h}-1}{h}$$ exists and hence defines some function $f(a)$. Next we need to show that this function is strictly increasing and its range is whole of $\mathbb{R}$. This would imply that there is a unique number $e$ such that the above limit $f(e)=1$. The proof of these statements is bit long and it is well explained in the blog post linked above.