Delta-Epsilon Proofs and Arbitrary Epsilon Choice

Date: 11/05/2002 at 11:52:32
From: Morris Waehlti
Subject: Do delta-epsilon proofs have to seem so arbitrary ? 

Dear Doctor Math,

The limit of f(x)=x^2 + 1 as x approaches 2 is 5.

Now a rigorous proof of this would seem to involve showing that 
for any choice of epsilon, the corresponding value for delta is 
min{1, epsilon/5}. Just what is this rigorous proof? How is delta = 
epsilon/5 obtained?

More generally, what is the nature of the special relationship between 
a function's behavior near a point, the choice of epsilon, and a 
corresponding "good" delta?

I hope that this is not too much to ask.

Very best regards,
Morris

Date: 11/05/2002 at 13:04:26
From: Doctor Peterson
Subject: Re: Do delta-epsilon proofs have to seem so arbitrary ? 

Hi, Morris. 

There are actually many ways you can choose delta; the choice of 1 in 
your answer is arbitrary, and any delta less than what you show will 
be satisfactory. There is no need to find the "best" delta, only one 
that will guarantee that the resulting error will be less than 
epsilon. In fact, there are entirely different ways to choose delta 
for this example, as you'll see in the second link below.

I think a general answer to your general question is that the bounds 
on the slope of the function in the neighborhood of the point under 
consideration will play a large role in determining what delta will 
look like.

We have several discussions of delta-epsilon proofs on our site, both 
explaining the general idea, and how to come up with deltas:

   Epsilon/Delta Definition of Limits
   http://mathforum.org/library/drmath/view/53738.html 

   A Limit Proof Using Estimation
   http://mathforum.org/library/drmath/view/53357.html 

   Formal Definition of a Limit
   http://mathforum.org/library/drmath/view/53403.html 

   Understanding the Need for Limits
   http://mathforum.org/library/drmath/view/53754.html 

I think these should answer your questions well. If you have any 
further questions, feel free to write back.

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

Date: 11/06/2002 at 10:36:03
From: Morris Waehlti
Subject: Do delta-epsilon proofs have to seem so arbitrary ? 

Dear Doctor Peterson,

Thank you for your comments and also for the library references, which 
I have already printed out but not yet read completely. 

Concerning this business about, "the bounds on the slope of the 
function in the neighborhood of the point under consideration will 
play a large role in determining what delta will look like," this is 
what I am thinking at this point: Let a particular situation be: 

   the limit of f(x) = x^2+1 as x approaches 2 is 5 

with epsilon = 1 and delta = 1/5. 

Now the neighborhood of the point 2 is (9/5,11/5) and the 
corresponding neighborhood of 5 is (4.24,5.84). The slopes of f 
corresponding to the endpoints of the neighborhood of 2 are 3.6 and 
4.4. So 3.6*1/5 = 0.72 and 5-4.24 = 0.76 are nearly the same. Also 
4.4*1/5 = 0.88, which is not far from 0.84. Has this got something to 
do with what it's about? Propagation of errors through functions?

Very best regards,
Morris

Date: 11/06/2002 at 12:03:35
From: Doctor Peterson
Subject: Re: Do delta-epsilon proofs have to seem so arbitrary ? 

Hi, Morris.

Yes, what you describe is what I had in mind. More precisely, the 
ratio of epsilon to delta will be, at least in "nice" cases like this, 
no more than the slope of the steepest chord from the limit point to 
points in the neighborhood. Because the parabola under consideration 
is smooth, with its slope either increasing or decreasing as you move 
away, this steepest chord will be at an endpoint of the interval:

                         R
     +-------------------+ f(2)+e = f(2+d)
     |                  /|
     |                 / |
     |                /  |
     |               / * |
     |              /    |e
     |             /     |
     |            / *    |
     |           /       |
     |         Q/*       |
     +---------+---------+ f(2)
     |      /* |         |
     |  / *    |         |
    P+ * - - - | - - - - | f(2-d)
     |         |         |
     |         |         |e
     |         |         |
     |         |         |
     |         |         |
     |         |         |
     +---------+---------+ f(2)-e
     |    d    |    d    |
    2-d        2        2+d

In this example, we can just find d (delta) for a given e (epsilon) 
by solving

    f(2+d) = f(2)+e

and then showing that, because of the way f curves, f(x) will be 
within e of f(2) whenever x is within d of 2. The ratio e/d is the 
slope of chord QR, which is between the slope of f at Q and the slope 
at R.

Other functions are not as simple, and we can't find the greatest 
possible delta so precisely; but we approximate this by making various 
simplifications. You could probably, in many cases, at least make a 
good guess at delta based on slope considerations, and then proceed to 
prove your conjecture algebraically. Of course, you realize that there 
is not just one delta we can use; any delta LESS than the "ideal" 
delta we found will satisfy the definition of a limit.

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

Date: 11/06/2002 at 13:34:58
From: Morris Waehlti
Subject: Thank you (Do delta-epsilon proofs have to seem so arbitrary 
? )

Thank you very much, Doctor Peterson, for your help.
Morris