# Dubious "proof" of $e^x$ derivative?

The proof to which I am referring is amply discussed here: Derivative of exponential function proof, but I remain unconvinced by the answers that pertain to the specific proof discovered by user1346994.

It all boils down to showing that $\lim_{h\to 0}\left({\dfrac{e^{h}-1}{h}}\right) = 1$

I find it highly deceiving to replace $e$ with $(1+h)^{1/h}$ in the expression in the limit.

https://math.stackexchange.com/a/671305/309566 is the answer in which I am most interested. But I'm still puzzled by the remark 'again by continuity' and the change of variables bit.

I suppose if my wish were modest, it would be a proof that follows the one stated but with clear justifications. Thanks and sorry if I sound confused!

There have been numerous questions on MSE dealing with same problem namely the derivative of $e^{x}$. Most of the time OP (as well as the people who answer the question) assumes some definition of symbol $e^{x}$ but forget to mention the definition of $e^{x}$ explicitly.

Note that a definition of $e$ alone is not sufficient to define the symbol $e^{x}$. The linked answer assumes many things almost all of which are very difficult to establish. In particular it assumes the following:

1) Definition of $x^{h}$ for all $h$ and $x > 0$ such that it is a continuous function of $h$.

2) Existence of limit $\lim_{t \to 0}(1 + t)^{1/t}$

3) Interchange of double limits in $t, h$ assuming continuity of a complicated function of two variables $t, h$.

Without justifying the above assumptions (or at least mentioning them explicitly and the fact that they need to be justified) the answer is a classic example of intellectual dishonesty. However most elementary textbooks on calculus are guilty of the same dishonesty so no one even thinks that this is actually a problem and studying calculus with non-rigorous proofs (or even no proofs at all) is almost a tradition.

A proper proof of derivative of $e^{x}$ must begin with definition of symbol $e^{x}$ and I have provided one such approach here.

My attempt:

For finite $h<1$ and integer $m$,

$$f(h):=\frac{e^h-1}h=\frac{\left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^h-1}h\\ =\lim_{m\to\infty}\frac{\left(1+\dfrac1m\right)^{mh}-1}h\\ =^* \lim_{m\to\infty}\frac{\left(1+\dfrac hm\right)^{m}-1}h\\ =\lim_{m\to\infty}1+\frac{m-1}{2m}h+\frac{(m-1)(m-2)}{3!m^2}h^2+\cdots\frac{h^m}{m^m}\\ =1+\frac h2+\frac{h^2}{3!}+\cdots$$ which is a convergent series as the terms are bounded by the powers of $h$, and the sum is bounded by

$$\frac1{1-h}.$$

Then by squeezing

$$\lim_{h\to0}f(h)=1.$$

By a change of variable $mh\to m$. We may also start from this line, as defining $e^h$.

Well the first question is if you believe that $\lim_{h \rightarrow \infty}(1+h)^{1/h} = e$. Whether you mathematically believe this really comes down to how you define $e$. One definition is quite simply as this limit. You then need to prove $\frac d{dx}e^x = e^x$, which seems to be what the author is doing.

I personally prefer to define $e$ as the exponential base having self-derivative, just because this is how people have been describing $e$ to me since I was a kid. Your proof is zero steps in this case.

I think the most common way to define $e$ is by defining $\ln$ first as the area (definite integral) under $1/x$, and defining $e$ so $\ln(e) = 1$. From here, proving the limit fact is a little harder, but getting the derivative is just a matter of chasing inverses around and using the fundamental theorem of calculus.

Anyway, say you assume this limit. You can get to $h \rightarrow 0$ instead of $\infty$ easily, actually. From here it's really just composition of (continuous!) functions: if $a(x) \rightarrow a_0$ then $b(a(x)) \rightarrow b(a_0)$. What's a little trickier is establishing the limit exists at all, so yes, there is work to make this proof complete, but it's not going in the wrong direction.

This may not convince you either and it is not a solution to the limit problem; another way of establishing the derivative of $e^x$.
Consider the inverse, $x=\ln y$, $y>0$ and differentiate both sides with respect to $x$ to get $1=\frac{1}{y}\frac{dy}{dx}$, which tells you $\frac{dy}{dx}=y$ where $y=e^x$. Then, we have $\frac{d}{dx}e^x=e^x$. We assume that we already know $\frac{d}{dx}\ln x=\frac{1}{x}$, $x>0$ here and there is an easier argument to establish that $\frac{d}{dx}\ln x=\lim\limits_{h\to0}\frac{\ln(x+h)-\ln(x)}{h}=\frac{1}{x}$, $x>0$.