I'm trying to find the easiest way to prove that $\lim_{h\to 0} \frac{e^h-1}{h}=1$. I found the following derivation, which seems the most straightforward: $$\lim_{h\to 0} \frac{e^h-1}{h} \overbrace{=}^* \lim_{h\to 0} \frac{((1+h)^\frac{1}{h})^h-1}{h}=\lim_{h\to 0}\frac{1+h-1}{h}=1$$ The step I'm curious about is the first one. I realize that $e=\lim_{h\to 0} (1+h)^\frac{1}{h}$, but what is it that fully justifies substituting this limit inside the outer limit? If I remember correctly sometimes making such substitutions inside limits can lead to false results?
Naively I would say that all I can say for sure is that $$\lim_{h\to 0} \frac{e^h-1}{h}=\lim_{h\to 0} \frac{(\lim_{k\to 0} (1+k)^\frac{1}{k})^h-1}{h}$$
Am I missing something trivial?
[Added: I define $e = \lim_{n\to\infty} (1+\frac{1}{n})^n$ and then $e^x$ is defined per the usual way of exponentiation of real numbers. Therefore the above limit is a stepping stone in proving that $(e^x)'= e^x$ and so we can't use this derivative (or its consequences, such as the Taylor series of $e^x$) in order to prove this limit, so that we avoid a cyclic argument.]
(Strictly speaking if my question is "is this step correct?" then the answers I was given here are clear: it's indeed incorrect, and it's unlikely that it's possible to salvage the "proof" by an elementary argument. The wider interpretation of the question, "is there a more motivated proof than the standard log/substitution one", remains unanswered.)
– Snaw Nov 28 '21 at 14:42