Epsilon/Delta Definition of LimitsDate: 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/ |
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