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Two Definitions of Limits, with Examples


Date: 05/11/98 at 05:23:25
From: Darryl Cain
Subject: Limit Theory

Dear Dr. Math,

I have seen many great articles from your homepage, and was wondering 
if you could help me with a definition. I need to know what Limit 
Theory is, and I can't find it anywhere.

I know it would be great to hear from you.

Thanks a lot,
Darryl Cain


Date: 05/11/98 at 15:09:27
From: Doctor Rob
Subject: Re: Limit Theory

I am not aware of a subject with the title "Limit Theory." One does 
study limits when one takes Calculus. There are several kinds of 
limits that are studied. One definition is like this:

   Given a function f(x) from the real numbers to the real numbers,
   and two real numbers a and L, we say that the limit of f(x) as x
   approaches a is equal to L, and write

      lim  f(x) = L
      x->a

   if, for every epsilon > 0, no matter how small, there exists a 
   delta > 0 such that 
   
      |x - a| < delta implies that |f(x) - L| < epsilon.

You can see that this has to do with values of the function f(x) being 
near L for all values of x near enough to a.

Example:

   lim  sin(x)/x = 1
   x->0

because for every epsilon > 0, no matter how small, we can put
delta = sqrt(6*epsilon). Then, since x > sin(x) > x - x^3/3! for 
x > 0, and since x - x^3/3! > sin(x) > x for x < 0, and sin(x) = x for 
x = 0, we have that:

   |sin(x) - x| <= |x^3/3!| = (|x|^3)/6

Then whenever |x| = |x - 0| < delta,

   |sin(x)/x - 1| <= (|x|^2)/6 < (delta^2)/6 <= epsilon

Another definition is like this:

   Given an infinite sequence of real numbers:

      {x(n): n = 1, 2, 3, ...}, 

and a real number L, we say that the limit of x(n) as n approaches 
infinity is equal to L, and write:

       lim       x(n) = L
   n->infinity

if, for every epsilon > 0, no matter how small, there is a natural 
number N > 0 such that n > N implies that |x(n) - L| < epsilon.

You can see that this has to do with the value of terms of the 
sequence being near L for all large enough values of n.

Example:

       lim     (n + 1)/(2*n + 1) = 1/2
   n->infinity

because, for every epsilon > 0, no matter how small, we can put
N > 1/(4*epsilon). Then whenever n > N > 1/(4*epsilon),

   epsilon > 1/(4*n) > 1/(4*n + 2)
                     = |(n + 1)/(2*n + 1) - 1/2| = |x(n) - L|

-Doctor Rob, The Math Forum
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