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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: 1,703
Registered: 1/8/12
Re: Does this imply that lim x --> oo f'(x) = 0?
Posted: May 24, 2013 3:28 AM
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On Thu, 23 May 2013, baclesback@gmail.com wrote:
> On Thursday, May 23, 2013 11:11:12 AM UTC-4, steine...@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.
>
> I think so; use the MVThm repeatedly. Starting in [0,1]:
>
> f(1)-f(0)=f'(c1)*1 , for c in (0,1)
>
> f(2)-f(1)=f'(c2)*1 ; c in (1,2)
> ...........
>
> f(n+1)-f(n)=f'(cn)*1


> Now, if f approaches a finite limit at oo , then , as n-->oo f(n+1)-f(n)
> =f'(cn) -->0 .


This can't be right because there's counter examles when f isn't monotone
and nowhere do you use monotonicity.

The problem is proving c_n -> 0 while all that you have is c_n in [0,1].
Even with that, the continuity of f' is needed to complete the proof.



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