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Epsilon/Delta Definition of Limits


Date: 08/26/99 at 01:18:12
From: Laurence Barton
Subject: Epsilon/delta definition of limits

I'm an older student trying to regain some lost ground in calculus. At 
this point in my course I'm having difficulty grasping the full 
concept of the epsilon/delta definition of limits. Graphically it 
makes complete sense. However, when you describe the same problem in 
numerical terms, some of it falls apart (for me, anyway.)

For example:

     Find the limit x^2 = 4 as x -> 2

     Solving using epsilon/delta limits:

          |x^2-4| < epsilon  whenever   0 < |x-2| < delta

[Everything looks okay up to this point.]

     Factored terms:

          |x-2||x+2| < epsilon

[This also seems to make sense, but now the text example seems to go 
into la-la land.]

          for all x in the interval (1,3)

[where did this interval come from? No mention of bounds was made 
until this point in the problem. No help to be found in previous 
examples either. Is this step supposed to be intuitive?]

* the example continues:

     we know that |x+2| < 5

[I guess they found the epsilon value by plugging in the upper bound 
of x into f(x)-L. I hate math books that don't leave at least a couple 
of bread crumbs to follow.]      

* finally the text solution:

     letting delta be a minimum of (epsilon/5) and 1.

[Epsilon/5 and 1?]

     it follows that whenever 0 < |x-2| < epsilon/5 we have 
     |x^2-4| = |x-2||x+2| < (epsilon/5)*5 = epsilon

[how did they get epsilon/5?]

Sincerely,
Laurence Barton


Date: 08/26/99 at 07:48:35
From: Doctor Jerry
Subject: Re: Epsilon/delta definition of limits

Hi Laurence,

>     Find the limit x^2 = 4 as x -> 2
>
>     Solving using epsilon/delta limits:
>
>          |x^2-4| < epsilon  whenever   0 < |x-2| < delta
>
>[Everything looks okay up to this point.]
>
>     Factored terms:
>
>          |x-2||x+2| < epsilon
>
>[This also seems to make sense. Now the text example seems to go into 
>la-la land.]

I suppose the author was thinking something like this: we want to 
make |x-2||x+2| small. We can control the |x-2| with the delta. That 
leaves |x+2|. Somehow we must control its size. So, if we begin by 
deciding that x is within 1 of its limit of 2. That makes x between 1 
and 3. This is arbitrary. One could say, let's require x to be between 
0 and 5. But one on either side of the limit 2 is natural. So, to 
begin with, we make delta less than 1. If this much is true, then x is 
in the interval (1,3) and so:

     |x-2||x+2| < |x-2|*5

Now all that remains is to put a SECOND requirement on delta. We want 
it so that:

     |x-2||x+2| < |x-2|*5 < delta*5 < epsilon

So, we choose delta to be the smaller of 1 and epsilon/5, so that both 
steps in our procedure hold.

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


Date: 08/28/99 at 16:46:29
From: Doctor Hans
Subject: Re: epsilon/delta definition of limits

Thanks for writing to Dr. Math. Let me try to add a few comments to 
your example.

You want to show that lim{x->2} x^2 = 4, or in the epsilon-delta 
terminology:

For all EPSILON > 0 there exixts a DELTA > 0 so that the following is 
true: If you choose x so that |x-2| < DELTA then you will have 
|x^2-4| < EPSILON.

So, in other words, given EPSILON > 0 your 'job' is to find DELTA > 0 
so that if you choose any number x in the interval (2-DELTA,2+DELTA), 
then this number will satisfy |x^2-4| < EPSILON.

In the example we will choose DELTA <= 1. Hence the interval (1,3). 
There is nothing special about the number 1 in this case; it is just 
to make our 'search' easier. We choose a smaller set in which to look 
for our DELTA.

     |x+2| < 5

This is obvious when we have decided only to look at x in (1,3).

     |x-2| < DELTA

This is true when we choose x in (2-DELTA,2+DELTA).

Note that (2-DELTA,2+DELTA) is contained in (1,3), so if we choose x 
in (2-DELTA,2+DELTA) we get, by the above:

     |x^2-4| = |x-2||x+2| < DELTA*5

If we choose DELTA so that DELTA*5 < EPSILON then we will have 
|x^2-4| < EPSILON (and we will be done).

But that is easy: choose DELTA < EPSILON/5. That is the reason why 
EPSILON/5 is chosen.

I hope that these comments make things clearer for you! If not, don't 
hesitate to write us back.

Sincerely,

- Doctor Hans, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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