Formal Definition of a Limit

Date: 10/17/2001 at 13:31:45
From: Rebecca Klusman
Subject: Calculus - the definiton of a limit

Could you please explain to me the formal definition of a limit. I 
need help specifically with finding a delta for a given epsilon and 
using the epsilon-delta definiton of a limit. The problems tell me 
to find the limit and then find delta is greater than 0 such that 
|f(x) - L| is less than 0.01 when ever 0 is less than |x - c| is less 
than delta. The other set of problems tells me to find the limit L and 
then use the epsilon-delta definition to prove that the limit is L.  

I have no idea how to do this at all - it's quite over my head. I am 
also taking this course by correspondence, so I have no one to help 
me. Thank you!

Date: 10/21/2001 at 15:27:22
From: Doctor Fenton
Subject: Re: Calculus - the definiton of a limit

Hi Rebecca,

Thanks for writing to Dr. Math. This isn't an easy concept to explain,
but see if the following helps.

The definition of a limit is a very sophisticated idea, and it's 
important to be sure you understand what you're trying to do.

lim f(x) = L  means that by taking a value of x very x->a close, but 
not equal, to a, the function value f(x) will be very close to L.  
We need an explicit test for this idea.

First of all, how do you TEST the statement "x is close to a"? Whether 
two numbers, say 5 and 6, are "close" is a matter of opinion, but 
given another number, say 5.5, we can definitely say that 5.5 is 
closer to 5 than 6 is. To decide if numbers are "close" we need a 
"standard of closeness," which can be any positive number. We will
say that two numbers p and q are close according to the standard c if 
|p-q|<c. For brevity, I will just say that p and q are c-close. So, if 
c = .01, 5.2 and 5 are NOT .01-close, but 5.005 and 5 are .01-close, 
because |5.2-5| = .2 > .01, while |5.005-5| = .005<.01 .

Think of the function f(x) as a "machine": a scientific or graphing 
calculator is a good example. You push a key indicating the function 
f, such as "sin" or "log," and type in a value of x, push "enter," and 
the display changes to the value f(x) of the function f at x. Imagine 
some machine generating random values of x that can be put into the 
machine r. You are the "input quality control inspector," you must 
decide whether to let the machine evaluate f(x), and you must base 
your decision solely on whether x passes an "input closeness" test, 
that is, whether x is close to a.

At the output end, an "output quality control inspector" will be 
judging your work. She will have her own standard of closeness, and 
she will test the output for closeness to L. If it passes, you get 
paid; if it fails, you will be fired.

So, you will have a positive number for an input closeness standard 
(the traditional name for the input closeness standard is the Greek 
letter delta, which I will write d), and the output inspector will 
have her own standard of closeness, traditionally called epsilon,
which I will write e.

There is one final consideration. Consider two functions f1 and f2, 
where

     f1(x)=x+1
and
     f2(x)=2x+1.

Clearly, if x is very close to 2, then f1(x) is very close to 3, while 
f2(x) is very close to 5, but we can be more precise. We can write

   |f1(x)-3| = |x+1-3|

             =|x-2| .

This equation says that the output closeness which the output 
inspector will measure is exactly equal to the input closeness, for 
the function f1.  

For the function f2, we have

   |f2(x)-7|=|2x+1-5|

            = |2x-4|

            = 2 |x-2|

which says that the "output closeness" is twice the input closeness.

In both cases, it is clear that there must be a relation between the 
input and output closeness standards. For f1, if the output inspector 
has a standard of closeness smaller than your input standard, some 
inputs will pass your test but fail the output test, and you will be 
fired. In fact, if the output inspector is out to get you fired, and 
if she knows your input standard, she can come in after you have left, 
run numbers that pass your input test through the machine until an 
f(x) comes out that is not exactly equal to L, and then choose her 
output standard so small that the output test will fail for that x.  
This test will only be fair if the output inspector publicly announces 
her output closeness standard, epsilon, before you have to choose your 
input standard, delta.

So, if you are the inspector for function f1, and the output inspector 
announces that epsilon = .01 will determine which outputs pass, you 
know you may simply take your input closeness standard, delta, also 
equal to .01, since you know, by your analysis of the machine, that 
the output closeness is equal to the input closeness, so if x passes 
the input test with delta = .01, the f1(x) will pass the output test 
|f1(x)-3|<.01 . Now, if the output inspector retires, and a new output
inspector arrives determined to improve the quality of output, so that 
from now on, epsilon will be .001, then you know you can keep your job 
if you pick delta = .001. If you then get promoted to becoming input 
inspector for f2, where the output inspector has a closeness standard 
of epsilon = .0001, then your analysis of the machine f2 shows that 
your input had better pass |x-2|<.00005, so you take delta = .00005.

The key is that no matter what output closeness standard, epsilon, 
is chosen, it is always POSSIBLE to find an input closeness standard, 
delta, that will guarantee that if the input x is delta-close to a,|x-
a|<d , then the output f(x) will be epsilon-close to L, |f(x)-L|<e.

That is, given any positive e, there exists a d such that any x 
(different from a) which satisfies |x-a|<d will produce a value f(x) 
that satisfies |f(x)-L|<e .

In a limit proof from the definition, your objective is to find a 
strategy for choosing d after e is given. For most of the problems 
usually encountered, you can find a relation between the input 
closeness and the output closeness of the form

   |f(x)-L|<= C|x-a| .

If you can find such an inequality, then it is easy to choose d when e 
is given. For a function satisfying such an inequality, the output 
closeness is no more than C times the input closeness, so you can 
guarantee that the output closeness 

   |f(x)-L| < e

if the quantity C|x-a|<e, since C|x-a| is known to be larger than 
|f(x)-L|.

Since C|x-a|<e if |x-a|<e/C, this suggests that we choose d as e/C. 
Then, if |x-a|<d = e/C, then

   |f(x)-L| <= C|x-a| < C[e/C] = e ,

and f(x) passes the output closeness test.

How do we find such an inequality for a given function? The functions 
f1 and f2 above show that it is easy (actually an equality) for linear 
functions f(x) = mx+b. What about f(x) = x^2 near a = 2?  We would 
expect that
   lim x^2 = 4
  x->2             .

Look at the output closeness:

   |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 .

If |x-2|<1, then -1 < x-2 < 1, or 1<x<3, and
 
       3 < x+2 < 5,

so that the factor

         |x+2| < 5.

Then, for x such that |x-2|<1,

   |x^2-4| < 5|x-2|.

Given e>0, choose d so that both

   |x-2| < 1  (to make the inequality true)

and so that

   |x-2|<e/5, to make |f(x)-L| < e.

I hope this is of some help to you. If you have further questions 
about this, please write back and I will try to explain further.

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