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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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Graham Cooper

Posts: 4,344
Registered: 5/20/10
Re: Does this imply that lim x --> oo f'(x) = 0?
Posted: May 24, 2013 5:51 PM
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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!

Any plot?

...


The un-differentiated fractal curve from above!

http://www.wolframalpha.com/input/?i=ln%28x%29*sin%28x*x%29%2Fx


Herc




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