Date: May 24, 2013 3:42 AM
Author: William Elliot
Subject: Re: Does this imply that lim x --> oo f'(x) = 0?
On Thu, 23 May 2013, email@example.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