On Thu, 23 May 2013 18:55:08 +0000, Bart Goddard wrote: > firstname.lastname@example.org 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).