How to proof that $\lim_{h \to 0}\frac{e^h-1}{h} = 1$ using the definition $e = \lim_{n \to \infty}(1+\frac{1}{n})^n$? In other words, how I can prove that these two definitions of $e$ is equal? I saw these two definitions while trying to find proofs for $\frac{d}{dx}e^x$ and $\frac{d}{dx}\ln x$; some use the former definition, and others used the latter, and I cannot find a proof that these two definitions are equal. So how do I prove this? Thank you!
 A: From $e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n$ easily follows $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$ for $x>0$.
If you prove that even $e=\lim\limits_{n\to-\infty}\left(1+\frac1n\right)^n$, then $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$ also for $x<0$.
Expand the expression inside the limit:
$$\left(1+\frac xn\right)^n=\sum_{k=0}^n\binom nk\frac{x^k}{n^k}=\sum_{k=0}^n\frac{n(n-1)\ldots(n-k+1)}{n^k}\frac{x^k}{k!}$$
For a fixed $k$ we have $\frac{n(n-1)\ldots(n-k+1)}{n^k}\to1$, so it's not hard to see $$\left(1+\frac xn\right)^n\to\sum\limits_{k=0}^\infty\frac{x^k}{k!}$$
using the fact that $\sum\limits_{k=0}^\infty\frac{x^k}{k!}<\infty$ and that $\frac{n(n-1)\ldots(n-k+1)}{n^k}<1$.
Your limit follows easily from this (in fact, we found the whole Taylor series for $e^x$).
A: I believe the problem here is that taking $e^x$ isn't properly defined. The definition $e = \lim_{n\to\infty} (1 + \frac{1}{n})^n$ is something I have always found dissatisfying. Here is an alternative approach that will also fulfill your desire for proofs of your derivative formulas.
Define log function $\mathrm{ln}: \mathbb{R}_+ \to \mathbb{R}$ as follows: $$\mathrm{ln}(x) = \int_{1}^{x} \frac{\mathrm{d}t}{t}$$
From this everything follows. We have immediately properties such as $\mathrm{ln}(ab) = \mathrm{ln}(a) + \mathrm{ln}(b)$. We also have by the Fundamental Theorem of Calculus that $\mathrm{ln}'(x) = \frac{1}{x}$. Note also that since $f:t\mapsto 1/t$ is strictly positive, $\mathrm{ln}$ is one-to-one. It is also not hard to show that $\mathrm{ln}$ is surjective. Thus, define $e$ as the unique positive real such that $\mathrm{ln}(e) = 1$. One can check that the natural log of your limit is one as well directly.
Additionally, define the exponential function $\exp(x)$ as the inverse of $\mathrm{ln}$. We can define then exponentiation of a real number $a^b = \exp(b \;\mathrm{ln}( a))$. Now this may seem circular: isn't exponentiation already defined? Only for $b \in \mathbb{Z}$. One can check that our definition of exponentiation confirms that $a^n = a*a*\ldots*a$ using the property that $\exp(a+b) = \exp(a)\exp(b)$ (proven by taking the inverse of both sides, essentially).
To take the derivative of $\exp(x)$, we use the chain rule:
$$1 = \frac{d}{dx}x = \frac{d}{dx} \mathrm{ln}(\exp(x)) = \frac{exp'(x)}{exp(x)}$$
Thus, $\exp'(x) = \exp(x)$.
One final note: Since $\mathrm{ln}(e) = 1$, we have that $\exp(x) = e^x$.
