TITLE: A natural discovery of $e$ as $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ - what is unrigorous here?
QUESTION [19 upvotes]: I had an interesting conversation with my teacher today, and we were discussing what the best foot forward for me might be. He asked me to differentiate $e^x$ from first principles, which I did, and then challenged me on:

Begin by defining $e$ as: $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

He wondered if I found this satisfactory, intuitively speaking. I said: "this is just one of the fundamental and historic definitions of $e$", and he asked me - "why?" And I just couldn't answer beyond repeating: "it's the definition". On that note he then showed a more natural exploration that leads to a discovery of this definition of $e$. He did declare this exploration to be "cheating", which is the focus of my question.

Take $a$ to be a positive real number. Then:$$\begin{align}\lim_{h\to0}\frac{a^{x+h}-a^h}{h}=a^x\lim_{h\to0}\frac{a^h-1}{h}\end{align}$$ And by exploration with graphs, one intuitively sees that the latter term, the derivative of $a^x$ at $0$, clearly should exist, and that there should exist some base $a$ for which this derivative is $1$. Exploring this possibility, define: $$(h_n)_{n\in\Bbb N}=\frac{1}{n}$$ And for each $h_n$, ask which $a$ satisfies: $$\frac{a^{h_n}-1}{h_n}=1$$ And naturally define a sequence $(a_n)$ that solves this equation: $$(a_n)_{n\in\Bbb N}=(1+h_n)^{1/h_n}$$ This leads to the idea that the mystery $a$ for which the derivative of $a^x$ at zero equals $1$ can be found by considering the limits of the sequence $a_n$, as $h_n$ goes to $0$. This finds: $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ And we call this number $e$. The exploration leads one to suspect that $e$ satisfies $\frac{d}{dx}e^x=1$.

However, he did mention that, in the interests of me studying real analysis, that this approach is "cheating" a little somehow, and that it is not a fully rigorous derivation of that limit. He hinted that I would see why this is shortly after properly studying real analysis.
I have seen a good deal of calculus and analysis around the Internet, in a very erratic way, so I have only some sense of why this approach is unrigorous. However, there are many many people on this forum who have properly studied real analysis, and I am asking for your help in explaining either what's wrong, or how to make it right.
My thoughts:

We have not shown that $a^x$ is sequentially convergent, nor have we shown that $(a_n)$ converges.
Showing that there exists $(a_n),(h_n)$ such that: $$\lim_{n\to\infty}h_n=0,\,\frac{a_n^{h_n}-1}{h_n}=1$$ Is not the same as showing that: $$\lim_{h\to0}\frac{(\lim_{n\to\infty}(a_n))^h-1}{h}=1$$

Any insight into $(1),(2)$ or anything else here would be greatly appreciated - especially any insight into how to make it fully rigorous! I just intuitively feel that $(2)$ is correct, but I cannot know for sure.

REPLY [4 votes]: The gaps in the argument that need to be filled with rigor are the following.

First, we need an airtight definition of just what $a^x$ means for at least a real number $a > 0$ and a real number $x$.
Second, we need to rigorously prove that the limit
$$\lim_{h \rightarrow 0} \frac{a^h - 1}{h}$$
exists.
Third, we need to prove that the limit
$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n$$
exists.
Finally, we need to insert the value of this limit, i.e. $e$ into the limit in (2) and show that the limit there is $1$, so that $\frac{d}{dx} e^x = e^x$.

Note here that because we do not have a different expression for the values of the limits above, we have to prove these limits exists as a fact in and of itself without finding their value. You may have done an $\epsilon$-$\delta$ proof before, but every such proof has involved that you already know the limit's value. That bit is important and that, together with (1), are where the real "real analysis" content lies here.