Proving Limits Formally
Date: 01/04/99 at 00:58:06 From: CNG Subject: What is a limit Dr. Math, I'm a high school student, and I'm trying to teach myself calculus. I've found many Web sites that attempt to teach this course, but I'm having serious trouble understanding them. I need to get an understanding in common English. My first problem is under the significance of limits. How do I determine what a limit is? Do I just gather a certain amount of information and take a guess? Do I keep increasing the number until I get the number that I'm trying to prove? For example, in trying to prove that .3333333... is equal to 1/3, I would do the Cauchy sequence: 3 33 333 3333 33333 333333 3333333 -- --- ---- ----- ------ ------- -------- 10 100 1000 10000 100000 1000000 10000000 I realize that with each addition of 3 onto the end, the number gets closer and closer to 1/3. Do I just say that 1/3 is the limit to this decimal? If so, then how do I know that I have enough information? Any help is appreciated. CNG
Date: 01/04/99 at 11:34:34 From: Doctor Rob Subject: Re: What is a limit Thanks for writing to Ask Dr. Math! You have found the way to guess what the limit is. Now you have to prove that your guess is the limit. To do that, you have to show that the sequence of fractions gets arbitrarily close to 1/3. Let the sequence be denoted by s(n), for n = 1, 2, 3, .... Now you have to show that the absolute value of the difference between s(n) and the limit L = 1/3 gets smaller than any pre-assigned value, that is that |s(n) - L| gets arbitrarily small as n gets large. Another way of saying this is that if I give you an epsilon > 0, no matter how small, you can find a N (depending on epsilon) such that for all n >= N, |s(n) - L| < epsilon That is exactly the wording of the definition of the limit of the sequence s(n) being L as n grows without bound. Now in your case, you have s(1) = 3/10 = 0.9/3 = (1-1/10)/3 s(1) - 1/3 = 3/10 - 1/3 = -1/30 s(2) = 3/10 + 3/100 = 33/100 = 0.99/3 = (1-1/100)/3 s(2) - 1/3 = 33/100 - 1/3 = -1/300 s(3) = 3/10 + 3/100 + 3/1000 = 333/1000 = 0.999/3 = (1-1/1000)/3 s(3) - 1/3 = 333/1000 - 1/3 = -1/3000 and so on. In fact, s(n) = (1-1/10^n)/3 |s(n) - 1/3| = 1/(3*10^n) Now this does get arbitrarily small as n gets large, because (note that <=> means if and only if) epsilon > 1/(3*10^n) <=> 10^n > 1/(3*epsilon) <=> n > log_10[1/(3*epsilon)] so we can pick any integer N >= log_10[1/(3*epsilon)], and apply the definition of a limit to prove that 1/3 is the limit of s(n) as n grows without bound. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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