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/
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