Approximation with 1-exponential

Question

How come that

$$\left(1-\frac{1}{x}\right)^x \approx e^{-1}\ ?$$

Is there a proof or something to understand this?

Answer 1

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 $,

$y = e^{-1}$

Answer 2

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.