Download How to..... DO AN EPSILON-DELTA ( ) PROOF BACKGROUND

........How to.....
DO AN EPSILON-DELTA (  ) PROOF
BACKGROUND:
Whenever there is a need for an    proof, you automatically know two things: (1) the proof is
not of the hand-wavy variety, but should be done with a high level of what mathematicians call
rigor...with utmost attention to logic, detailed justification of each step, and an ultimate tie-back to
the basic definitions of the concepts involved, and (2) the proof will generally be aimed at validating
a limit, which in turn might be used to decide whether a function is continuous somewhere or that
sort of thing.
The motivation for the definition of limit is very intuitive. What would a reasonable concept of
the limit of a function look like, anyway? You want to capture the idea that if a variable is "pretty
close" to a particular value, say x  c, then some formula based on that variable, say fx, is going to
be "pretty close" to some other value which we will call the limit. We haven’t said what "pretty
close" means yet. And we actually don’t care what the value of the formula is when we plug the
particular value x  c into it. This may seem odd...you might say "if you have a chance to evaluate
fx at the particular value c of the variable, why not just do it and call that the limit?" The thing is,
we are more interested in behavior near c rather than exactly at it. It often happens that the trend of
2
the formula near c is all you have (think in terms of the fx  x  1 example at x  c  1, where
x1
f1 isn’t even defined because of division by zero, but the trend near x  1 is very predictable), or
the trend wildly disagrees with the evaluation of the formula at the particular point fc for some
reason. So we ask: "If there is a value L that the formula fx seems to trend towards when we move
x closer and closer to the particular value c that we have chosen?" In other words, given a tolerance
(allowable variation) on fx around L, can we find a corresponding tolerance on x around (but not
at!) c so that given tolerance on fx is not exceeded? If we can do that, then we can sensibly say that
the limit of fx as x approaches c is L. The given or desired tolerance around L will be the epsilon
and the tolerance we are required to find will be the delta.
Calculus thrived for 200 years without the epsilon-delta definition of limit being formally
available because its leading practitioners had such good intuition that they were probably using it
subconsciously. It is traditional to give the idea of taking a limit a bit of dynamic character by using
the word "approach", although the situation is really quite static. The usual supposition is that
epsilon is small, but that is unnecessary as well. The crucial philosophical point is that no matter
what epsilon is given...big, medium, or small...a corresponding delta must be found with the requisite
properties.
DEFINITION:
More verbal: Given the real function fx defined on the open set E, we say that the limit of
fx as x approaches c  E is L if, for every   0, there exists a corresponding   0, which
generally depends on both  and c, such that whenever x is no further away from c than  (but
doesn’t equal c), then we can be assured that fx is no further away than  from L.
More mathematical: Given f : E  , and c  E  , where E is open, then lim xc fx  L
iff   0, , c  0 such that 0  |x  c|  , c  |fx  L|  .
METHOD:
The proof won’t construct the limit for you...you have to have it either given or be able to make a
good guess what L is. Once you have a candidate limit L and the function fx whose limit L is
supposed to be at the particular point c, the drill goes like this. Start out by writing down "Fix
  0". Now you need to fiddle with algebra to find a  that whenever |x  c|  , it forces
|fx  L|  . Usually  turns out to be some mildly complicated function of . Remember that you
will be working with absolute values, so there are some rules peculiar to them that must be observed.
You may need to account for the point c in the specification of . I’ll give you some model proofs
below which illustrate this. After you do the algebra, you present the  that is supposed to work for
that fixed  at the given point c, and it may not hurt to work back thru it to demonstrate it is valid.
MODEL PROOFS:
1) Let’s show that lim x0 3 x  0.
Proof: Fix   0. We want , 0 such that |x  0|  , 0 will force | 3 x  0|  . I’ve written
the zeroes just to show where they go. Of course this all is equivalent to |x|  , 0 must imply
| 3 x |  . Now cubing both sides of the latter inequality, we get |x|   3 . I’ve used the fact that
| 3 x |  3 |x| . OK...look at where |x| has to be relative to , and where it is relative to , 0. If we set
, 0   3 , then whenever |x|  , 0   3 we have | 3 x |  , as required by the limit definition,
and we are done. Implicitly I have also used the fact the for any real numbers a  b  0, it is true
that a 3  b 3 . You can visualize this easily by thinking of the graph for the cubing function.
2) OK...a little more complicated this time...let’s show lim x8 3 x  2.
Proof: Fix   0. We want , 8 such that |x  8|  , 8 will force | 3 x  2|  . We could
cube | 3 x  2| and try to work with that as we did above, but the expansion of the cube will be messy.
And we will have the sum of terms inside of the absolute value bars. Remember, absolute values are
simple in products or quotients, but less so in sums or differences. Here are the rules, by the way:
|a|
.....|a  b|  |a|  |b|.....|a  b|  ||a|  |b||. Alternatively, is there a hidden
|ab|  |a||b|..... a 
b
|b|
factorization in x  8? Yes there is, and we can force the cube root to appear using the identity
a 3  b 3  a  ba 2  ab  b 2 . In our case, a 3  x and b 3  8. So
|x  8|  | 3 x  2 3 x  2  2 3 x  4|  | 3 x  2| . | 3 x  2  2 3 x  4|. Do you see that keeping
|x  8|
less than ? They are equal! So our
| 3 x  2| less than  is the same as keeping
2
3
| x   2 3 x  4|
|x  8|
job is to ensure
  by keeping |x  8|  , 8. Equivalently, we want to keep
| 3 x  2  2 3 x  4|
|x  8|    | 3 x  2  2 3 x  4|. If we could put a lower bound on | 3 x  2  2 3 x  4|, let’s call it B
for the moment, then our job is to keep |x  8|  B, which will be achieved if we set , 8  B.
So let’s find B. Actually we don’t "find" it so much as force it to be a certain value by a technique
that is common to these kinds of proofs. What if I say that , 8  8 no matter what? We are trying
to find some formula for , 8 in terms of , and for a particular value of , it could be that
, 8  8. I want to prevent that, so I will take my formula for , 8 as long as it doesn’t exceed 8,
and if it does, then I will ignore it and say , 8  8. OK, why did I choose 8? If |x  8|  8, then
0  x  16, and I can use 0 in the expression | 3 x  2  2 3 x  4| to find B, the lower bound. Notice
how simple putting x  0 into the expression is...I get 4 immediately. So now if , 8  8, then
|x  8|  4, so I finally set , 8  min8, 4, and we are done. Note that if we are given   2,
then 8 cuts in as , 8, and if   2 then , 8  4. Either way, we are covered.
© 2008 Thomas Beatty