```Date: Jan 27, 2013 12:31 PM
Author: Charles Hottel
Subject: Re: Limit Problem

"William Elliot" <marsh@panix.com> wrote in message news:Pine.NEB.4.64.1301261746060.3500@panix3.panix.com...> On Sat, 26 Jan 2013, Charles Hottel wrote:>>> I am having a problem following an example in my book.>>>> Prove  lim(x->c) 1/x  = 1/c, c not equal zero>>> Assume c /= 0, |x - c| < s, s < c>> -s < x - c < s> c - s < x < c + s>> 1/(c + s) < 1/x < 1/(c - s)> 1/(c + s) - 1/c < 1/x - 1/c < 1/(c - s) - 1/c = (c - c + s)/c(c - s)>> -s/c(c - s) < 1/x - 1/c < s/c(c - s)> |1/x - 1/c| < s/c(c - s)>> (c^2 - cs)r = s> Let s = rc^2 / (1 + cr)>> s/c(c - s) = [rc^2 / (1 + cr)] / c(c - rc^2 / (1 + cr))> rc^2 / c(c + rc^2 - rc^2) = r>> Given r > 0, take s as above to show> |x - c| < s implies |1/x - 1/c| < r.>> Be sure to make s small enough so that s < c.>>> So 0 < | x-c| < delta,  implies |1/x - 1/c| < epsilon>>>> |1/x - 1/c| = | (c-x) / {xc}| = 1/|x| * 1/|c| * (x-c) < epsilon>>>> Factor 1/|x| is troublesome if x is near zero, so we bound it to keep it>> away from zero.>>>> So |c| = |c - x + x| <= |c-x| + |x| and this imples |x| >= |c| - |x-c|>>>> I think I understand everything up to this point, but  not the next >> steps,>> which are>>>> If we choose delta <= |c|/2 we succeed in making |x| >= |c| / 2.>> Finally if we require delta <= [(epsilon) * (c**2)} / 2 then>>>> [1/|x| * 1/|c| *  |x-c|]  <  [1 / (|c|/2)]  *  [1/|c|]  *  [((epsilon) *>> (c**2)) / 2] = epsilon>>>> How did they know to choose delta <= |c|/2?>>>> How does that lead to |x| > |c|/2 implies 1/|x| < 1/(|c|/2) ?>>>> I did not sleep well last night and I feel I must be missing something>> that would be obvious if my head was clearer.  Thanks for any help.>>>>>>>>Thanks, I had sort of figured it out on my own but you post makes it clearer.
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