```Date: Jan 27, 2013 4:22 AM
Author: William Elliot
Subject: Re: Limit Problem

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 < sc - s < x < c + s1/(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 = sLet 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) = rGiven 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.> > > >
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