Limit Proofs

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
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