Date: 08/23/98 at 15:39:41 From: Amanda Donovan Subject: Calculus III My problem is that I quite simply don't understand where all things come from when doing a proof for a limit. My professor writes things on the board, such as the statement: If |f(x) - L| < E, then 0 < |x-a| < delta .... I don't understand the process. Is there a systematic way to do these proofs? The specific question I'm working on is: show that the limit as x approaches 0 of 1/x-1 = -1. I come out with a number and no delta in my answer. Any help you can give will be greatly appreciated. Sincerely, Amanda Donovan
Date: 08/24/98 at 08:04:54 From: Doctor Jerry Subject: Re: Calculus III Hi Amanda, First, you must mean 1/(x-1), right? I'll give an argument to show that the limit of 1/(x-1) as x->0 is -1, using the definition of limit you outlined above. We want to force |1/(x-1) - (-1)| < E by controlling the size of |x-0|. Let's work with |1/(x-1) + 1| < E, in an attempt to get |x-0|: |1/(x-1) - (-1)| = |1/(x-1) + 1| = |x|/|x-1| This will be small when |x| is small, provided we can control 1/|x-1|. Let's first decide to make delta < 1/2. If this is true, then when |x| < delta, -1/2 < x < 1/2. Thus, x is no closer to 1 than 1/2, that is, |x-1| > 1/2. Draw a small figure if this isn't clear. If |x-1| > 1/2, then 1/|x-1| < 2. Now back to |x|/|x-1|. Since 1/|x-1| < 2, we can say that: |1/(x-1) - (-1)| = |1/(x-1) + 1| = |x|/|x-1| < 2|x| To make this less than E, we need 2|x| < E, which means that |x| < E/2. So, we choose delta this way: For any E > 0, let delta be the smaller of E/2 and 1/2. Now, all of the above arguments work. - Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.