Slope definition of $e$ A common definition of $e$ is given as
$$e = \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}$$
which can be proven to be equivalent to
$$e=\lim_{h\rightarrow 0}\ \left(1+h\right)^{\frac{1}{h}}$$
The most practical use of $e$ in elementary calculus is however given as
$$1=\lim_{h\rightarrow 0}\frac{e^h - 1}{h}$$
which is used as a statement the slope of $e^x$ at $x=0$ is $1$ allowing one to prove that $\frac{d}{dx}e^x=e^x$. It appears trivial to prove that the two limits given above are equivalent, but I cannot seem to make any progress without making some illegal limit operations. I suspect the problem is deeper than it appears (I suspect the trouble is that although we have defined $e$, we have not actually said anything about what $e^x$ is). How does one rigorously proceed from the given definition of $e$ to the slope limit?
 A: I'm not sure if you have multiple questions here but I'll try to answer all of them.
For deriving:
$$1=\lim_{h\rightarrow 0}\frac{e^h - 1}{h}$$
We start with the definition of a derivative:
$$f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$
If we let $f(x)=a^{x}$, where $a \in \mathbb{R}$ then,
$$f'(x) = \lim_{h \rightarrow 0} \frac{a^{x+h} - a^{x}}{h}$$
$$f'(x) =\lim_{h \rightarrow 0} \frac{a^{x}(a^{h} - 1)}{h}$$
$$f'(x) =a^{x} \left( \lim_{h \rightarrow 0} \frac{a^{h} - 1}{h} \right)$$
If we let $a=e$, since $e \in \mathbb{R}$ we have that
$$f'(x) = e^{x} \left( \lim_{h \rightarrow 0} \frac{e^{h} -1}{h} \right)$$
Using l'hopital's rule and that $e^{h}$ is continuous at $h=0$ we can say that,
$$f'(x)= e^{x} \ln(e)$$
So, if we evaluate it at $x=0$ we have,
$$f'(0)=e^{0}=1$$
So that is how you obtain it one way.
Another way would go as follows:
Let $f(x)=\ln(x)$, then $f'(x) = \frac{1}{x}$, so $f'(1)=1$ and then we say,
$$1= \lim_{h \rightarrow 0} \frac{f(1+h) - f(1)}{h}$$
Because I like x's
$$1 = \lim_{x \rightarrow 0} \frac{f(1+x) - f(1)}{x}$$
$$1 = \lim_{x \rightarrow 0} \frac{\ln(1+x) - \ln(1)}{x}$$
$$1 = \lim_{x \rightarrow 0} \frac{1}{x} \ln(1+x)$$
$$1 = \lim_{x \rightarrow 0} \ln(1+x)^{\frac{1}{x}}$$
Raise both sides by $e$ and we get,
$$e^{1} = e^\left({\lim_{x \rightarrow 0} \ln(1+x)^{\frac{1}{x}}}\right)$$
$$e = \lim_{x \rightarrow 0} \text{ }e^\left({\ln(1+x)^{\frac{1}{x}}}\right)$$
$$e = \lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}}$$
And there you have your second formulation. For the first one, just let $x = \frac{1}{n}$ and you receive,
$$e = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^{n}$$
Hopefully that helps! Let me know if I made a mistake somewhere.
A: It is easy to go from the first limit to the second and back by using the substitution 
$$z=\ln(1+h) \,,$$
Note that this is equivalent to $h= e^{z}-1$.
and the continuity of the $\ln()$.
$$e=\lim_{h\rightarrow 0}\ \left(1+h\right)^{\frac{1}{h}} \Leftrightarrow $$
$$1=\lim_{h\rightarrow 0}\ {\frac{\ln \left(1+h\right)}{h}} \Leftrightarrow$$
$$1=\lim_{z\rightarrow 0}\ {\frac{z}{e^z-1}} $$
Writing down a formal proof is easy now, you just have to be carefull filling it the details. You need to use the continuity of the logarithm function, the fact that $z=\ln(1+h)$ is a bijection from a  neighborhood of zero to a neighborhood of zero; and most importantly that for each of the two implications $e$ and hence $\ln$ are defined in a different way. Thus, you cannot really work with if and only if, because then $\ln$ makes no sense. But you can do each implication separately easely...
