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Topic: Does this imply that lim x --> oo f'(x) = 0?
Replies: 3   Last Post: May 24, 2013 1:34 AM

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James Waldby

Posts: 358
Registered: 1/27/11
Re: Does this imply that lim x --> oo f'(x) = 0?
Posted: May 23, 2013 5:09 PM
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On Thu, 23 May 2013 18:55:08 +0000, Bart Goddard wrote:
> 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.


> Something like f(x) = sum_1^infinity arctan(2^n ( x-2^n) )/2^n should
>
> work. f'(x) is a sum of terms like 1/(1 + (2^n x -2^(2n))^2.
> f'(2^n)=1 plus some small positive terms. But f'(2^n+2^(n-1)) should
> be pretty close to zero.


As another example, f(x) = ln(x)*sin(x*x)/x goes to zero and
d(f(x))/dx goes to +/-oo as x goes to +oo, since it is dominated
by ln(x)*cos(x*x).

<http://www.wolframalpha.com/input/?i=d%28ln%28x%29*sin%28x*x%29%2Fx%29%2Fdx>
shows a graph for small x of d(f(x))/dx = d/dx((log(x) sin(x x))/x)
= (2 x^2 log(x) cos(x^2)-(log(x)-1) sin(x^2))/x^2.

--
jiw



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