Is it possible to prove the derivative of $e^x$ is $e^x$ using the limit definition of $e$ without using binomial expansion? I am teaching my students about the derivative of $e^x$. I have walked through what is in our textbook and they have all happily believed me, but I would like to have a better explanation for why the derivative is $e^x$. I know several other proofs exist like those that define $e$ using the slope at $0$ and those that use the natural log, but I'd like my proof to closely follow the method used in the textbook.  That is why this question and this question have not answered all of my questions. Here is what the textbook says.

I understand everything until they decide to let $e^{\Delta x}\approx 1 + \Delta x$. My question is why do they not directly use the limit definition of $e$. Is it because they can't without introducing students to binomial expansion?
I put $e^{x}$ outside the limit and then tried to substitute the definition of $e$. But I think I made a mistake. Here is my work:
$$e^{ x} \lim_{\Delta x\to\ 0} \frac{(\lim_{\Delta x\to\ 0}( 1+ \Delta x)^{\frac{1}{\Delta x}}) ^{\Delta x}-1}{\Delta x}$$
Using limit properties I know I can rewrite this as:
$$e^{x} \lim_{\Delta x\to\ 0} \frac{(\lim_{\Delta x\to\ 0}( 1+ \Delta x))-1}{\Delta x}$$
This is where I get confused/stuck.
If I resolve the inner limit first, I get:
$$e^{x} \lim_{\Delta x\to\ 0} \frac{1-1}{\Delta x}$$ $$e^{x}\lim_{\Delta x\to\ 0} \frac{0}{\Delta x}$$ $$e^{x} \times 0 = 0$$
Is there a property of limits that would allow me to "get rid" of that inner limit?
 A: In fact, by limits properties, you can put the definition of $e$ inside, just not in the way you did.
Given that $e= \lim_{\Delta x \to 0} (1+\Delta x)^{\frac{1}{\Delta x}},$ then
\begin{align*}
f'(x) &= e^{ x} \lim_{\Delta x\to\ 0} \frac{(( 1+ \Delta x)^{\frac{1}{\Delta x}}) ^{\Delta x}-1}{\Delta x}\\
&=e^x \lim_{\Delta x\to\ 0} \frac{(( 1+ \Delta x)-1)}{\Delta x}\\
&=e^x
\end{align*}
Another approach if we want to be more rigorous:
Its clear that, if $x \to \infty,$ we have
$$\left(1 + \frac{1}{x}\right)^x \leq e \leq \left(1 + \frac{1}{x}\right)^{x+1}$$
Replacing with $1/x= h$, then for $h \to 0^+$
$$(1+h)^{1/h} \leq e \implies 1 \leq \frac{e^h-1}{h},$$
and because of the RHS inequality:
\begin{align*}
1 \leq \frac{e^h-1}{h} \leq \frac{(1+h)^{1+h}-1}{h} \leq \frac{(1+h)(1+h^2)-1}{h} = \frac{h+h^2+h^3}{h} = 1 + h + h^2.
\end{align*}
We conclude by squeeze theorem the value of the limit (consider $h=\Delta x$).
Note that in the last inequality we used the fact that $(1+h)^h \leq 1+h^2$ for $0<h<1$. This is given by Bernoulli inequality.
A: You made a mistake in your work:
$$\frac{e^x\left(e^{\Delta x}-1\right)}{\Delta x}$$
is not the same as
$$\frac{e^x\left(\lim_{\Delta x\to0}\left(\left(1+\Delta x\right)^{\frac{1}{\Delta x}}\right)^{\Delta x}-1\right)}{\Delta x}.$$
Using $\Delta x$ in the definition of $e$ is the cause of the confusion. Let's use a different name, let's say $h$, in the definition of $e$. We then get
$$\frac{e^x\left(e^{\Delta x}-1\right)}{\Delta x}=\frac{e^x\left(\left(\lim_{h\to 0}\left(1+h\right)^{\frac{1}{h}}\right)^{\Delta x}-1\right)}{\Delta x}=\lim_{h\to 0}\frac{e^x\left(\left(1+h\right)^{\frac{\Delta x}{h}}-1\right)}{\Delta x}.$$
By definition of $\lim_{h\to 0}$ it's the common limit of all the sequences where we replace $h$ by $h_n$ with $h_n\to0$ as $n\to\infty$. Therefore
$$\frac{e^x\left(e^{\Delta x}-1\right)}{\Delta x}=\lim_{n\to \infty}\frac{e^x\left(\left(1+\frac{\Delta x}{n}\right)^{n}-1\right)}{\Delta x},$$
where we used $h_n=n^{-1}\Delta x$. In other words we get
$$\frac{e^x\left(e^{\Delta x}-1\right)}{\Delta x}-e^x=\lim_{n\to \infty}\frac{e^x\left(\left(1+\frac{\Delta x}{n}\right)^{n}-1-\Delta x\right)}{\Delta x}.$$
You can estimate using the binomial formula (or using induction similar to Bernoulli's inequality or ...) for example (for $|\Delta x|<1$)
$$\left\lvert\frac{\left(\left(1+\frac{\Delta x}{n}\right)^{n}-1-\Delta x\right)}{\Delta x}\right\rvert\leq |\Delta x|.$$
Therefore putting it together (for $|\Delta x|<1$)
$$\left\lvert\frac{e^x\left(e^{\Delta x}-1\right)}{\Delta x}-e^x\right\rvert\leq e^x|\Delta x|.$$
A: This is because your inner limit is not actually a limit, it is an approximation. The limit is used to make an approximaton when $\Delta x$. You are not taking the limit (at first), you are stopping at some point where you can use the approximation because the inner and the outer limits have the same variable, thus you can't separate their going to 0. Let us illustrate what went wrong in your reason with the following example. Say I am trying to evaluate $\lim_{x\rightarrow\infty} 2x/x$, I know that $2x$ is trivially approximated by $2x$ for smaller $x$. But $2 = \lim_{x\rightarrow 0} 2x/x \neq \lim_{x\rightarrow 0} (\lim_{x\rightarrow 0} 2x)/x = \lim_{x\rightarrow 0} 0/x$.  What the $\approx$ symbol is saying is that when $\Delta x$ is small, $f(\Delta x)$ ressembles another  more desirable function, which will be used instead.
A way to think about it is that since
$$e = \lim_{a\rightarrow 0} (1+a)^{1/a}$$
for small enough $a$, $e$ is close to $(1+a)^{1/a}$ leading to $f(a) = e^a$ being close to $g(a)=((1+a)^{1/a})^a = 1 + a$ (for how close you'd have to use Taylor Expansion). Now when $\Delta x \rightarrow 0$, $\Delta x$ is small enough that $e^{\Delta x}$ is as close as we wish to $1 + \Delta x$. The two funstions become interchangeable, and you can replace $f(\Delta x) = e^{\Delta x}$ by $g(\Delta x) = (1+\Delta x)$ so that, for small $\Delta x$, $\frac{f(\Delta x) - 1}{\Delta x} \approx \frac{g(\Delta x) - 1}{\Delta x}$. It then follows that
$$ \lim_{\Delta x \rightarrow 0} \frac{f(\Delta x) - 1}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{g(\Delta x) - 1}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{1 + \Delta x - 1}{\Delta x} = 1$$
The "for small values of $\Delta x$", "close" really are shortcuts for epsilon delta reasoning but this is also what the "$\approx$" symbol does...
