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Re: Limit Problem
Posted:
Jan 27, 2013 4:22 AM


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. > > > >



