Proving that $\lim_{h \to 0}\frac{e^h-1}{h}=1$

Question

I understand that this question has been asked many times, but I'm wondering if the approach I am using can be made rigorous. Say we define $e$ as $\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$. In general, $a^b$ is defined through continuity. How can we prove that $\frac{d}{dx}(e^x)=e^x$? Here is what I have tried so far: \begin{align} \frac{d}{dx}(e^x) &= \lim_{h \to 0}\frac{e^xe^h-e^x}{h} \\ &= e^x \lim_{h \to 0}\frac{e^h-1}{h} \, . \end{align} Then, to evaluate $\lim_{h \to 0}\frac{e^h-1}{h}$ we make the substitution $h=1/n$: \begin{align} \lim_{h \to 0}\frac{e^h-1}{h} &= \lim_{h \to 0}\frac{\left(\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n\right)^h-1}{h} \\ &= \lim_{h \to 0}\frac{\left(\lim_{h \to 0}\left(1+h\right)^{1/h}\right)^h-1}{h} \\ &= \lim_{h \to 0}\frac{(1+h)-1}{h} \\ &= 1 \, . \end{align} Unfortunately, I feel like some of the steps used here were a little hand-wavey. I'm not sure the way I 'absorbed' one of the limits into the other is actually justified, and nor do I see a way of proving that $$ \left(\lim_{h \to 0}(1+h)^{1/h}\right)^h=\lim_{h \to 0}(1+h)^1=1 \, . $$ So was the method I used rigorous, and if not, can it be amended?

Answer 1

Short answer: define $e$ another way first, then prove the definitions' equivalence.

Long answer, very carefully worded to avoid circularity:

For $a>0$, define $a^x$ in the obvious way for $x\in\Bbb Q$, then impose continuity. You can similarly show $\frac{d}{dx}a^x=a^x\ln a$ with $\ln a:=\lim_{h\to0}\frac{a^h-1}{h}$. Since $(a^b)^x=a^{bx}$, by the chain rule $\ln(a^b)=b\ln a$.

Provided $a\ne1$, we can take $e:=a^{1/\ln a}$ so $\ln e=1$, and $\frac{d}{dx}e^{kx}=ke^{kx}$ for all $k\in\Bbb R$. Since $e$ is therefore unique, $\ln a:=\log_ea$.

If $y^\prime=y$ then $(y/e^x)^\prime=\frac{(y^\prime-y)e^x}{e^{2x}}=0$ so $y\propto e^x.$ But$$\frac{d}{dx}\lim_{n\to\infty}(1+x/n)^n=\lim_{n\to\infty}\frac{d}{dx}(1+x/n)^n=\lim_{n\to\infty}(1+x/n)^{n-1}=\lim_{n\to\infty}(1+x/n)^n$$(the first $=$ exchanges two limits, which I think you can justify with this), so $\lim_{n\to\infty}(1+x/n)^n=e^x$ (the proportionality constant follows from the case $x-0$). Now take $x=1$.

Answer 2

This is just $e'(0)$, by the definition of the derivative.