Show that lim logx = 0 as x->1 by using the e-d format.
Can use inequalities:
logx<=x-1 for all x>0
logx>=2(x-1) for all x in [.5,1]
Definition of e-d:
lim f(x) = b as x->a
if for each e>0 there exists d>0 such that |f(x)-b|<e whenever 0<|x-a|<d
For any epsilon > 0:
Let delta = min { 0.5, epsilon/2 }
By definition, 0 < delta <= 0.5.
Therefore, any x that satisfies |x - 1| < delta, is somewhere inside the interval [0.5, 1.5]. (Because delta <= 0.5.)
Let x be a number that satisfies |x - 1| < delta.
If x = 1: |log(x)| = 0, therefore |log(x)| < epsilon.
If x > 1: |log(x)| = log(x) <= x - 1 < delta <= epsilon/2 < epsilon
If x < 1: (0.5 <= x < 1): |log(x)| = -log(x) <= -2(x-1) <= 2|x-1| < 2*epsilon/2 = epsilon
In all 3 cases: |log(x)| < epsilon.
Therefore, by Weierstrass' epsilon-delta definition, lim (x->1) log(x) = 0. Q.E.D.
Yeah, sadly on my final I only showed x=1 and x>1 and completely forgot about the x<1 case. Professor was not pleased.
In which class were you being asked to use the delta-epsilon proof to find a limit?