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Re: Limit Problem
Posted:
Jan 27, 2013 12:31 PM


"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 <  xc < delta, implies 1/x  1/c < epsilon >> >> 1/x  1/c =  (cx) / {xc} = 1/x * 1/c * (xc) < 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 <= cx + x and this imples x >= c  xc >> >> 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 * xc] < [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.



