Intriguing Limit

Date: 04/30/97 at 19:33:57
From: oliver
Subject: (1-1/x)^x

I am curious to see what chance you have of winning a contest if you 
have a 1 in 10 chance of winning each time you play and you play 10 
times.  I computed this as follows: chance = 1-.9^10 = .65...  What 
would it be for a 1 in 20 chance if you play 20 times? How about a 1 
in 100 chance when you play 100 times? And so on.

The function then looks like y = 1-(1-1/x)^x, where x is the chance 
you have of winning and the number of times you play. So if x = 100 
you have a 1 in 100 chance of winning each time you play and you play 
100 times.  y is then the chance that you will win at least once.  

What is the limit of this as x->infinity? When I graph it, it looks 
asymptotic at about .632..., but the function looks to me like it has 
a limit of 0.  (1/x goes to 0 and 1 to any power is one.) I asked a 
professor about this, and he told me that (1-1/x)^x has a special 
limit, namely 1/e. This then gives the expected result.

My question is: Why is lim (1-1/x)^x = 1/e?

Thank you,
Oliver Dale

Date: 05/01/97 at 09:01:05
From: Doctor Jerry
Subject: Re: (1-1/x)^x

Hi Oliver,

There are several different reasons "why" this limit has this 
particular value. I'll give one answer below, based on a calculation 
using l'Hopital's Rule, from calculus. If you're asking why is this 
limit the number 2.71828..., the base of the natural logarithms, the 
answer is that this limit comes up in working with exponentials and 
logarithms and was long ago given the name e, after Euler. Usually, 
it comes up in the form (1+1/n)^n, which approaches e as n->oo.

Let y = (1-1/x)^x.  Take natural logs of both sides.

ln(y) = x*ln(1-1/x).  As x->oo, this has the form of oo*0 and so must 
be rearranged.  We have:

ln(y) = ln(1-1/x)/(1/x)

Now the form is 0/0 and so l'Hopital's Rule is applicable. After 
differentiating numerator and denominator separately (according to 
l'Hopital's Rule) and simplifying (I'll use lim to mean limit as x->
oo), we have:

lim ln(y) = lim  (-1)/(1-1/x) = -1.

So ln(y) -> -1.  This means that y->e^(-1).

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/