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Topic: Does this imply that lim x --> oo f'(x) = 0?
Replies: 18   Last Post: May 26, 2013 1:28 AM

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 William Elliot Posts: 2,637 Registered: 1/8/12
Re: Does this imply that lim x --> oo f'(x) = 0?
Posted: May 24, 2013 3:42 AM

On Thu, 23 May 2013, steinerartur@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

Date Subject Author
5/23/13 steinerartur@gmail.com
5/23/13 Robin Chapman
5/23/13 David C. Ullrich
5/23/13 Bart Goddard
5/23/13 steinerartur@gmail.com
5/23/13 William Elliot
5/24/13 steinerartur@gmail.com
5/24/13 Bacle H
5/24/13 William Elliot
5/24/13 Bacle H
5/24/13 William Elliot
5/24/13 Graham Cooper
5/24/13 Graham Cooper
5/25/13 steinerartur@gmail.com
5/25/13 Graham Cooper
5/26/13 Bacle H
5/26/13 Bacle H
5/24/13 William Elliot
5/24/13 Graham Cooper