Evaluating Nested Limits

Question

In my attempt to prove $\frac d{dx}\left(e^x\right)=e^x$, I arrived at the following step: $$\frac d{dx}\left(e^x\right)=e^x\lim_{h\to0}\frac{e^h-1}{h}$$ At this point, I substituted in $$e=\lim_{n\to\infty}{\left(1+\frac 1n\right)^n}$$ to arrive at $$\frac d{dx}\left(e^x\right)=e^x\lim_{h\to0}\frac{\lim_{n\to\infty}{\left(1+\frac 1n\right)^{nh}}-1}{h}$$ Then, I let $h=\frac 1n$ to get $$\frac d{dx}\left(e^x\right)=e^x\lim_{h\to0}\frac{(1+h)^{h\left(\frac 1h\right)}-1}{h}$$ at which point a lot of things simplify and the limit is clearly equal to one. My question concerns the $h=\frac 1n$ step. While I see why this might be true, I don't really have a good conceptual or mathematical grasp of why this works, or whether this is even valid at all. Can the nested limit just be removed that way and $n$ replaced, or is there something else at work here?

Answer 1

It is not legitimate to let $h=1/n$ since $h$ and $n$ are independent parameters. We can still proceed along any of the following ways forward.

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METHODOLOGY $1$:

In THIS ANSWER, I showed that $\left(1+\frac xn\right)^n$ converges uniformly to $e^x$ on any bounded interval of the real line.

Therefore, the function $\frac{d}{dx}\left(1+\frac xn\right)^n=\left(1+\frac xn\right)^{n-1}$ converges uniformly to $e^x$ on any bounded interval of the real line.

Hence, $\frac{de^x}{dx}=e^x$ for all $x\in\mathbb{R}$.

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METHODOLOGY $2$:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x}$$

for $x<1$. Therefore, we can write for $0<h<1$

$$1\le \frac{e^h-1}h\le \frac{1}{1-h}$$

whence application of the squeeze theorem yields the coveted limit

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

A parallel development for $h<0$ results in the same conclusion. And we are done.

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METHODOLOGY $3$:

In THIS ANSWER, I discuss the Moore-Osgood Theorem, which permits the interchange of limits if one of the two limits converges pointwise and the other uniformly.

It can be shown that $\lim_{h\to 0}\frac{\left(1+\frac1n\right)^{nh}-1}{h}=n\log\left(1+\frac1n\right)$ and the convergence is uniform for $n\in \mathbb{N}$.

Hence, the Moore-Osgood Theorem guarantees that

$$\begin{align} \lim_{h\to 0}\lim_{n\to \infty}\frac{\left(1+\frac1n\right)^{nh}-1}{h}&=\lim_{n\to \infty}\lim_{h\to 0}\frac{\left(1+\frac1n\right)^{nh}-1}{h}\\\\ &=\lim_{n\to \infty} n\log\left(1+\frac1n\right)\\\\ &=1 \end{align}$$

as expected again!