Proving existence and uniqueness of the number $e$ as the exponential function base with derivative 1 at $x=0$ with $\lim_{h\to 0} \frac{e^h-1}{h}=1$.

Question

I was just revisiting the definition of the number $e$ in Section 4.3 of Zbigniew Nitecki's Calculus Deconstructed. I'll start by saying I really like his ground-up development of the exponential and logarithmic functions in Section 3.6, establishing that the exponential function defined over the rationals for any positive real base has a removable discontinuity at each irrational number, which allows it to be well-defined over the reals as the limits of sequences over the domain of rationals.

In Section 4.3, where he develops the derivatives of exponentials and logarithms, for any $b>0$ we are given the definition of the function $\varphi_b$ on $(-\infty,0)\cup(0,\infty)$ as $\varphi_b(x) = \frac{b^x-1}{x},$ so that $\left.\frac{\mathrm d b^x}{\mathrm d x}\right\rvert_{x=0} = \lim_{x\to 0} \varphi_b(x),$ if this limit exists, and the following:

Lemma 4.3.1. For any $b > 0$ and any $a\in(-\infty,\infty),$ $\left.\frac{\mathrm d b^x}{\mathrm d x}\right\rvert_{x=a} = b^a \left(\left.\frac{\mathrm d b^x}{\mathrm d x}\right\rvert_{x=0}\right)= b^a \lim_{x\to 0} \varphi_b(x),$ provided that the limit exists.

Remark 4.3.2. The function $\varphi_b$ is positive if $b>1$ and negative if $0<b<1$.

Lemma 4.3.3. For $b>1$ and $n$ any nonzero integer, $\frac{b^{n+1}-1}{b^n-1} \geq \frac{n+1}{n}.$ The inequality is strict for $n\neq -1.$

Lemma 4.3.4 For every positive $b\neq 1,$ $\varphi_b$ is strictly increasing on its natural domain $(-\infty,0)\cup(0,\infty)$.

Corollary 4.3.5. For each $b>0,$ the limit $\psi(b)=\left.\frac{\mathrm d b^x}{\mathrm d x}\right\rvert_{x=0}=\lim_{x\to 0} \varphi_b(x)$ exists, and hence $f(x)=b^x$ is differentiable everywhere, with derivative at $x=a$ given by $\frac{\mathrm d}{\mathrm d x}[b^x] = \psi(b)\cdot b^x.$

Lemma 4.3.6.

1. For $b>1,$ $\psi(b)>0,$ while for $0<b<1,$ $\psi(b) < 0.$

2. For any $b>0$ and $r\in\mathbb R,$ $\psi(b^r) = r\psi(b).$ In particular, there exists a value of $b$ for which $\psi(b)=1.$

Definition 4.3.7. $e$ is the unique number satisfying $\lim_{h\to 0} \frac{e^h-1}{h}=1.$

I noticed that while Lemma 4.3.6 establishes the existence of a base for the exponential function with derivative $1$ at $x=0$, using the lemma with any base greater than $1,$ for example taking the base of $2,$ so we have from part 1 that $\psi(2) > 0,$ and so setting $r = 1/\psi(2)$ we then have from part 2 that $\psi(2^r)=r\psi(2)=1,$ how do we know this is the only value for which $\psi$ is $1$?

Answer 1

Letting $e$ be a number satisfying the equation in Definition 4.3.7, we can use the facts proven in section 3.6 about the continuity, monotonicity, and unboundedness of exponential functions to draw on these answers to duplicate questions:

• One option would be to prove that $\psi$ is monotonically increasing, as done here, to conclude there could only be one such value for $e$. This answer also shows how we can solve for the value of $e$ satisfying this equation to be $\lim_{n\to\infty}(1 + \frac{1}{n})^n.$
• An even faster option, as used here, is to recognize that every other positive real number $x$ can be written as $e^c$ for some $c \neq 1$, so then $\lim_{h \to 0}\dfrac{x^h - 1}{h} = \lim_{h\to 0}\dfrac{e^{ch} - 1}{ch}c = c \neq 1$, and thus any other $x$ could not satisfy Definition 4.3.7.

See this answer and the other answers to this question for more on connecting the limit and infinite series definitions of $e$.

Answer 2

Fix some $b>1$, for instance $b=2$.

• Let us first prove that $$f: \Bbb R\to\Bbb R_+\quad r\mapsto b^r$$ is a bijection. By corollary 4.3.5 and the first part of lemma 4.3.6, $f$ is continuous and strictly increasing. Moreover, $\lim_\infty f=\lim_{n\to\infty}b^n=\infty$ hence $\lim_{-\infty}f=\lim_\infty\frac1f=0$.

• Now, from $$\forall r\in\Bbb R\quad\psi(f(r)) = r\psi(b)$$ we derive $$\forall s\in\Bbb R_+\quad\psi(s) = f^{-1}(s)\psi(b)$$ hence $$\psi(s)=1\iff s=f\left(\frac1{\psi(b)}\right).$$

Answer 3

Hint:

In general, $(b^x)'=(e^{x\ln b})'=e^{x\ln b}\cdot \ln b=\ln b\cdot b^x.$

So evaluation at zero gives $\ln b\cdot b^0=\ln b\cdot 1=\ln b.$

If you want this to be $1$ we need $b=e.$