Epsilon-Delta ProofsDate: 09/28/2004 at 08:36:29 From: Tharshan Subject: Epsilon-Delta Proof I have two questions on epsilon-delta proofs, one from your site and one found in the book "Calculus 8th edition: Varberg, Purcell, Rigdon". I hope you can help me with them. First, an example from the book: lim 1/x = 1/c x->c Proof: We need 0 < |x - c| < delta so that |1/x - 1/c| < epsilon. After doing some calculations, it comes to: (1/|x|)(1/|c|)(|x - c|)<epsilon Now they say in the book, "The factor 1/|x| is troublesome, especially if x is near 0. To that end note that: |c| = |c + x - x| =< |c - x| + |x| |x| >= |c| - |x - c| Thus if we choose delta =< |c|/2, we succeed in making |x| => |c|/2. Finally, if we also require delta =< (epsilon)(c)^2/2 then (1/|x|)(1/|c|)(|x - c|)<(1/|c|/2)(1/|c|)(epsilon)(c)^2/2 = epsilon" My question is that I don't understand all those things in the quotes, especially how they got delta =< |c|/2. So I hope you guys can clear that up for me, please. My second question concerns this link from your archives: http://mathforum.org/library/drmath/view/53403.html You have given a proof regarding lim x->c for x^2 = 4. I don't understand why the following is done: |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. What does it mean to let |x+2| < 5? Is this the "upper bound" that I've been hearing about? What does upper bound mean? I hope you guys can help me clear up all these doubtful questions. Thank you very much for this service that is helping so many out there. Date: 09/28/2004 at 13:28:34 From: Doctor Peterson Subject: Re: Epsilon-Delta Proof Hi, Tharshan. The idea in your first question is that you have to find conditions so that (1/|x|)(1/|c|)(|x - c|) < epsilon when |x - c| is small enough, but 1/|x| can get very large if x is small. If c is near zero, then, the expression on the left can get large if you aren't careful to keep x away from 0. So we look at a number line: |=======| <-- keep x in here, <------+---+---+------> and it will stay away from 0 0 c/2 c In order to keep x away from 0, we just have to keep it close enough to c, namely closer than |c/2|. That's the basic idea. The details are a matter of proving that this will do the trick. The triangle inequality is used to prove that if |x-c| < |c|/2, then |x| > |c|/2, which is obvious from my number line picture! Then we find that under those conditions, (1/|x|)(1/|c|)(|x-c|) < (1/|c|/2)(1/|c|)delta = 2/c^2 delta so that we will be within epsilon as long as delta is chosen small enough so that the right side is less than epsilon. Is that enough to guide you through what they're doing? Turning to your second question, it's similar to the other. If |x-2| < delta, then |f(x) - L| < |x+2| delta so that as long as we can be sure that |x+2| is kept small enough (within bounds, that is, so that it never exceeds some fixed number), we can promise that the right side will be less than epsilon. Again, we look at a number line: |=======| <-- keep x within here, and <------+---+---+---+----> x+2 will be no greater than 5 0 1 2 3 We've arbitrarily chosen the numbers here--we are talking about keeping x close to 2 in the first place; we arbitrarily choose to make sure that delta is no greater than 1, and see that under those conditions, x+2 will be no greater than 3+2, which is 5. That puts a bound on |x+2|; that is, we know the highest it could possibly get. So as long as |x-2| < delta < 1 we know that |x+2| < 5 and therefore |f(x)-L| < |x+2| |x-2| < 5 delta So by requiring delta to be less than both 1 and epsilon/5, we know that we can keep |f(x)-L| less than epsilon. The algebra just proves that this all really works. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 09/28/2004 at 22:12:59 From: Tharshan Subject: Thank you (Epsilon-Delta Proof) Thank you very much Dr. Peterson for your explanations, they very much helped me. Thanks again. Date: 09/28/2004 at 22:41:11 From: Doctor Peterson Subject: Re: Thank you (Epsilon-Delta Proof) Hi, Tharshan. Thanks for letting me know this was what you needed. I think it is common for mathematicians to give all the algebraic details but have trouble explaining the motivation behind what they do, which is what makes math "alive" and interesting. Maybe it's because we are taught too well to be rigorous, and the heuristic thinking that comes first seems like a dirty little secret, even though it's really the foundation of much of what we do! The fact that you have no trouble with the rigorous part, but needed help with the behind-the-scenes thinking, speaks well for you. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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