Date: May 24, 2013 5:51 PM
Author: Graham Cooper
Subject: Re: Does this imply that lim x --> oo f'(x) = 0?
On May 24, 5:42 pm, William Elliot <ma...@panix.com> wrote:
> On Thu, 23 May 2013, steinerar...@gmail.com wrote:
> > Suppose f:[0, oo) --> R is increasing, differentiable and has a finite
> > limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0? I guess
> > not, but couldn't find a counter example.
> No, it doesn't.
> For x in [0,1], let f(x) = 0.
> For x in [n, n+1/2], let f(x) = 1 - 1/n.
> For x in [n+1/2, n+1],
> . . let f(x) = 1 - 1/n + (x - n + 1/2)(1/n - 1/(n+1)).
> To assure f' exists, round the corners at n, n + 1/2, for all n in N.
> To make f strictly increasing, slope slightly upwards the horizontal
> portion. gtest
OK, so you approach the limit in smaller and smaller bumps!
Another self-similar fractal coastline but increasing!
The un-differentiated fractal curve from above!