Approximation with 1-exponential How come that 
$$\left(1-\frac{1}{x}\right)^x \approx e^{-1}\ ?$$
Is there a proof or something to understand this?
 A: The natural (i.e. base-$e$) exponential function is its own derivative.  That means its growth rate is equal to its present size.  Let's say $x$ is one million.  Being an exponential function, the function is multiplied by the same amount every time a millionth of a unit of time passes.  The size now is $1$; the size one millionth of a unit of time ago is the present growth rate, which is $1$, multiplied by the time, one millionth.  Therefore one millionth of a unit of time ago, the size was $1-(1/x)$.
Siince it's a base-$e$ exponential function, the size one full unit of time ago is $e^{-1}$.
Every time you go one millionth of a unit of time into the past, you multiply the size by the same amount, and as we saw, that amount is $1-(1/x)$.  To get to one unit of time in the past, you have to multiply by that number $x$ times, in our example one million times.  Therefore, when you multipy $x$ times by $1-(1/x)$, you get about $e^{-1}$.  That's not exact because a millionth of a unit of time can be further subdivided, giving you a still closer approximation to $e^{-1}$.
This is of course not a rigorous proof.  Often heuristic arguments are more enlightening.
A: let   $y = \left(1-\frac{1}{x}\right)^x \ $, taking log both side, we get,
${\log(y) = x(\log(1 - {1\over{x}}))}$, Now by taylor expansion of log we get,
$\log(y) = x(\ {-1\over{x}} -  ({-1\over{x}})^{2}.{1\over{2}} +\  ...)  $
$\log(y) = (\ {-1} -  x({-1\over{x}})^{2}.{1\over{2}} +\  ...)  $,  take limit both side,
$\lim_{x\to\infty}\log(y) =  -1  $,
$y = e^{-1}$
