Associated Topics || Dr. Math Home || Search Dr. Math

### 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/
```
Associated Topics:
High School Analysis
High School Calculus

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search