Am I allowed to define $e$ through algebraic means using this limit?

Question

By using the formal definition of a derivative

$$f'(x) = \lim \limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

one can get

$$\frac{d}{dx}[e^x] = e^x \times \lim \limits_{h \to 0} \frac{e^h-1}{h}$$

which implies

$$ \lim \limits_{h \to 0} \frac{e^h-1}{h} = 1$$

If we remove the limit but remember that $h$ is a really small number one can then do the following

$$ \begin{align*} \frac{e^h-1}{h} = 1 \\ \\ e^h-1=h\\ \\ e^h=h+1\\ \\ e=(h+1&)^\frac{1}{h} \end{align*}$$

Apply the limit to get

$$e=\lim \limits_{h \to 0} (h+1)^\frac{1}{h}$$

Which is correctly one of the representations/definitions of $e$. Then one can set $n = \frac{1}{h}$ and get the common

$$e=\lim \limits_{n \to \infty} (\frac{1}{n}+1)^n$$

But it doesn't feel right however. It feels like I cheated. So am I allowed to actually derive $e$ like this? Is it "formal" in other words.

Answer 1

No.

In one of your first steps, you at best have $$ \frac{e^h-1}h\approx 1.$$ And applying $(\cdot)^{\frac1h}$ might principally make the small-ish error much larger.

E.g., we know that $\lim_{n\to\infty}\sqrt[n]n= 1$. Arguing like you tried, ...

If we remove the limit but remember that $n$ is a really big integer, one can then do the following $$\sqrt[n]n=1,$$ $$n=1^n,$$ $$n=1.$$

This is obviously wrong.