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Re: Does this imply that lim x > oo f'(x) = 0?
Posted:
May 23, 2013 5:09 PM


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 ( x2^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^(n1)) 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



