# 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?

-
This is true in the limit, i.e. as $x\to \infty$, because $$\lim_{x\to\infty}\left(1+\frac{y}{x}\right)^x=e^y$$ – Alex Becker Sep 14 '12 at 17:28
ah it's a common limit, thank you! I didn't realize it. Make it an answer and i'll accept it – John Smith Sep 14 '12 at 17:33
@JohnPell : I think you may find the answer I posted more edifying, if you digest it carefully. – Michael Hardy Sep 14 '12 at 17:44
See this link; math.stackexchange.com/a/195093/8581. Brian's answer tells you more about what you are searching for. – Babak S. Sep 14 '12 at 18:42

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,

$\lim_{x\to\infty}\log(y) = -1$,

$y = e^{-1}$

-

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.

-
I guess the question really is what is the definition of e$^x$ and what is a consequence of the definition. Is e$^x$ defined as the limit of (1+x/n)$^n$ and d/dx (e^$x$) =e$x$ a consequence or is it the other way around. I think this is a matter of convention. As I remember while learning calculus the limit was the definition and the property of the derivative was a consequence. – Michael Chernick Sep 14 '12 at 17:51
The norms of deductive logic say you can take any of several characterizations to be the definition and then deduce the others from it. But in addition to the norms of deductive logic, there's the question of which characterization best explains to the students why one introduces such a number as $e$ in the first place, and what's "natural" about it. One way to do that is to write an easy proof that $(d/dx) a^x = (a^x\cdot\text{constant})$ and then talk about what $a$ has to be in order for the "constant" to be $1$. If it's done that way, then the answer above makes sense. – Michael Hardy Sep 15 '12 at 2:20
I agree. I think that I basically said the same thing just pointing out that I learned it the other way. Aside from that point which is just an observation your approach is fine as a way to present an answer to the OP. – Michael Chernick Sep 15 '12 at 3:48