TITLE: Particular definition of e
QUESTION [5 upvotes]: Show that, $$e=\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x $$
Is the same number that satisfies, 
$$\lim_{h\to0} \frac{e^{h}-1}{h} = 1 \tag{*}$$
You don't have to do that if it's too cumbersome, but (*) is used to find the derivatives of real exponential functions. I know that definitions are not to be proven but I'm looking for some derivation or intution for the claim that (*) is a legitimate definition for e, and furthermore, a proof that limit in (*) exists.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

I like the following procedure because it shows how to choose the 'best' logaritm base $\ds{\mrm{a}}$: By 'best' I mean the simplest $\ds{\log}$-derivative expression.

\begin{align}
\totald{\log_{\,\mrm{a}}\pars{x}}{x} & \stackrel{\mrm{def.}}{\equiv}
\lim_{h \to 0}{\log_{\,\mrm{a}}\pars{x + h} - \log_{\,\mrm{a}}\pars{x} \over h} =
\lim_{h \to 0}{1 \over h}\log_{\,\mrm{a}}\pars{1 + {h \over x}}
\\[5mm] & =
{1 \over x}\log_{\,\mrm{a}}\pars{\lim_{h \to 0}\bracks{1 + {h \over x}}^{x/h}}
\end{align}
The 'simplest $\ds{\,\mrm{a}}$-choice' is given by:
$$
\mrm{a} = \lim_{h \to 0}\pars{1 + {h \over x}}^{x/h} =
\lim_{n \to \pm\infty}\pars{1 + {1 \over n}}^{n} = \color{#f00}{\expo{}}
\quad\mbox{which yields}\quad
\totald{\log_{\,\mrm{a}}\pars{x}}{x} = {1 \over x}
$$