Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Epsilon-Delta Proofs

Date: 09/28/2004 at 08:36:29
From: Tharshan
Subject: Epsilon-Delta Proof

I have two questions on epsilon-delta proofs, one from your site and
one found in the book "Calculus 8th edition: Varberg, Purcell,
Rigdon".  I hope you can help me with them.  First, an example from 
the book:

lim 1/x = 1/c
x->c

Proof: 

We need 0 < |x - c| < delta so that |1/x - 1/c| < epsilon.  After
doing some calculations, it comes to:

  (1/|x|)(1/|c|)(|x - c|)<epsilon

Now they say in the book, "The factor 1/|x| is troublesome, especially
if x is near 0. To that end note that:

  |c| = |c + x - x| =< |c - x| + |x|
  |x| >= |c| - |x - c|  

Thus if we choose delta =< |c|/2, we succeed in making |x| => |c|/2. 
Finally, if we also require delta =< (epsilon)(c)^2/2 then

  (1/|x|)(1/|c|)(|x - c|)<(1/|c|/2)(1/|c|)(epsilon)(c)^2/2 = epsilon"

My question is that I don't understand all those things in the quotes,
especially how they got delta =< |c|/2.  So I hope you guys can clear
that up for me, please.


My second question concerns this link from your archives:

  http://mathforum.org/library/drmath/view/53403.html 

You have given a proof regarding lim x->c for x^2 = 4. I don't 
understand why the following is done:

  |f(x)-L| = |x^2 - 4|
           = |x+2||x-2|

  Here, the output closeness is a variable multiple of the input
  closeness. However, we are only interested in x's that are close to 
  2, so we can make a preliminary restriction to consider only those
  x's that are at least within 1 unit of 2. That is, we first require 
  |x-2| < 1.

What does it mean to let |x+2| < 5? Is this the "upper bound" that 
I've been hearing about?  What does upper bound mean?  I hope you guys
can help me clear up all these doubtful questions.  Thank you very
much for this service that is helping so many out there.



Date: 09/28/2004 at 13:28:34
From: Doctor Peterson
Subject: Re: Epsilon-Delta Proof

Hi, Tharshan.

The idea in your first question is that you have to find conditions so
that

  (1/|x|)(1/|c|)(|x - c|) < epsilon

when |x - c| is small enough, but 1/|x| can get very large if x is 
small.  If c is near zero, then, the expression on the left can get 
large if you aren't careful to keep x away from 0.

So we look at a number line:

             |=======|      <-- keep x in here, 
  <------+---+---+------>       and it will stay away from 0
         0  c/2  c

In order to keep x away from 0, we just have to keep it close enough 
to c, namely closer than |c/2|.  That's the basic idea.  The details 
are a matter of proving that this will do the trick.  The triangle 
inequality is used to prove that if |x-c| < |c|/2, then |x| > |c|/2, 
which is obvious from my number line picture!  Then we find that under 
those conditions,

  (1/|x|)(1/|c|)(|x-c|) < (1/|c|/2)(1/|c|)delta = 2/c^2 delta

so that we will be within epsilon as long as delta is chosen small 
enough so that the right side is less than epsilon.

Is that enough to guide you through what they're doing?


Turning to your second question, it's similar to the other.  If 
|x-2| < delta, then

  |f(x) - L| < |x+2| delta

so that as long as we can be sure that |x+2| is kept small enough 
(within bounds, that is, so that it never exceeds some fixed number), 
we can promise that the right side will be less than epsilon.  Again, 
we look at a number line:

             |=======|      <-- keep x within here, and
  <------+---+---+---+---->     x+2 will be no greater than 5
         0   1   2   3

We've arbitrarily chosen the numbers here--we are talking about 
keeping x close to 2 in the first place; we arbitrarily choose to make 
sure that delta is no greater than 1, and see that under those 
conditions, x+2 will be no greater than 3+2, which is 5.  That puts a 
bound on |x+2|; that is, we know the highest it could possibly get.  
So as long as

  |x-2| < delta < 1

we know that

  |x+2| < 5

and therefore

  |f(x)-L| < |x+2| |x-2| < 5 delta

So by requiring delta to be less than both 1 and epsilon/5, we know 
that we can keep |f(x)-L| less than epsilon.  The algebra just proves 
that this all really works.

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 09/28/2004 at 22:12:59
From: Tharshan
Subject: Thank you (Epsilon-Delta Proof)

Thank you very much Dr. Peterson for your explanations, they very much 
helped me. Thanks again.



Date: 09/28/2004 at 22:41:11
From: Doctor Peterson
Subject: Re: Thank you (Epsilon-Delta Proof)

Hi, Tharshan.

Thanks for letting me know this was what you needed.  I think it is
common for mathematicians to give all the algebraic details but have
trouble explaining the motivation behind what they do, which is what
makes math "alive" and interesting.  Maybe it's because we are taught
too well to be rigorous, and the heuristic thinking that comes first
seems like a dirty little secret, even though it's really the 
foundation of much of what we do!  The fact that you have no trouble
with the rigorous part, but needed help with the behind-the-scenes
thinking, speaks well for you.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
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

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/