As to your comment:
Consider the differential equation
$y - \left( {1 + \frac{x}{n}} \right)y' = 0$
It's solution is clearly $y_n={\left( {1 + \frac{x}{n}} \right)^n}$
If we let $n \to \infty$ "in the equation" one gets
$y - y' = 0$
One should expect that the solution to this is precisely
$\lim_{n \to \infty} y_n =y=\lim_{n \to \infty} \left(1+\frac x n \right)^n := e^x$
Also note $\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{{xn}}} \right)^{xn}} = \mathop {\lim }\limits_{n \to \infty } {\left[ {{{\left( {1 + \frac{1}{n}} \right)}^n}} \right]^x}$
My approach is the following:
I have as a definition of $\log x$ the following:
$\log x :=\lim_{k \to 0} \frac{x^k-1}{k}$
Another one would be
$\log x = \int_1^x \frac{dt}t$ Any ways, the importance here is that one can define $e$ to be the unique number such that
$\log e =1$
so that by definition
$\log e =\lim_{k \to 0} \frac{e^k-1}{k}=1$
From another path, we can define $e^x$ as the inverse of the logarithm. Since
$(\log x)'=\frac 1 x$
the inverse derivative theorem tells us
$(e^x)'=\frac{1}{(\log y)'}$
where $y=e^x$
$(e^x)'=\frac{1}{(1/y)}$
$(e^x)'=y=e^x$
The looking at the difference quotient, one sees that by definition one needs
$\mathop {\lim }\limits_{h \to 0} \frac{{{e^{x + h}} - {e^x}}}{h} = {e^x}\mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - 1}}{h} = {e^x}$
so that the limit of the expression is $1$. One can also retrieve from the definition of the logarithm that
$\eqalign{ & \frac{x}{{x + 1}} <\log \left( {1 + x} \right) < x \cr & \frac{1}{{x + 1}} < \frac{{\log \left( {1 + x} \right)}}{x} <1 \cr} $
Thus
$\mathop {\lim }\limits_{x \to 0} \frac{{\log \left( {1 + x} \right)}}{x} = 1$
a change of variables $e^h-1=x$ gives the result you state. In general, we have to go back to the definition of $e^x$. If one defines ${e^x} = 1 + x + \frac{{{x^2}}}{2} + \cdots $
Then
$\frac{{{e^x} - 1}}{x} = 1 + \frac{x}{2} + \cdots $
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \left( {1 + \frac{x}{2} + \cdots } \right) = 1$
from the defintion we just chose.