Particular definition of e 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. 
 A: For the forward implication, given that $e = \lim_{x \to \infty} (1 + 1/x)^x$, we have for $n \in \mathbb{N}$
$$e = \lim_{n \to \infty}\left(1 + \frac{1}{n} \right)^n = \lim_{n \to \infty}\left(1 + \frac{1}{n} \right)^n \left(1 + \frac{1}{n} \right) =  \lim_{n \to \infty}\left(1 + \frac{1}{n} \right)^{n+1}. $$
It s straightforward to show using Bernoulli's inequality that $(1+1/n)^n$ is increasing and $(1 + 1/n)^{n+1}$ is decreasing.  
Hence,
$$\left(1 + \frac{1}{n} \right)^n \leqslant e \leqslant \left(1 + \frac{1}{n} \right)^{n+1},$$
and
$$1 \leqslant \frac{e^{1/n}-1}{1/n} \leqslant n\left[\left(1 + \frac{1}{n} \right)\left(1 + \frac{1}{n} \right)^{1/n}- 1\right] =  (n+1)\left(1 + \frac{1}{n} \right)^{1/n}- n .$$
Using Bernoulli's inequality we have 
$$(n+1)\left(1 + \frac{1}{n} \right)^{1/n} - n  \leqslant (n+1) \left(1 + \frac{1}{n}\frac{1}{n}\right) - n  = 1 + \frac{1}{n} + \frac{1}{n^2}.$$
Thus,
$$1 \leqslant  \frac{e^{1/n}-1}{1/n} \leqslant 1 + \frac{1}{n} + \frac{1}{n^2}.$$
Applying the squeeze theorem we see that
$$\tag{*}\lim_{n \to \infty} \frac{e^{1/n}-1}{1/n} = 1.$$
Continuity and monotonicity of the exponential function $x \mapsto e^x$ can be used to show that (*) implies 
$$\lim_{h \to 0} \frac{e^{h}-1}{h} = 1.$$
A: Consider a sequence $a_n$ that satisfies
$$\frac{(a_n)^{1/n}-1}{1/n}=1$$
It then follows that
$$a_n=\left(1+\frac1n\right)^n$$
And as $n\to\infty$...
(a bit of monotone magic may be in order)
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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:
$$
\mrm{a} = \lim_{h \to 0}\pars{1 + {h \over x}}^{x/h} =
\lim_{n \to \pm\infty}\pars{1 + {1 \over n}}^{n} = \color{#f00}{\expo{}}
\quad\mbox{which yields}\quad
\totald{\log_{\,\mrm{a}}\pars{x}}{x} = {1 \over x}
$$
