On Friday, May 24, 2013 3:28:09 AM UTC-4, William Elliot wrote: > On Thu, 23 May 2013, bacail.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.
How about this: with the same lay out as before: f(n+1)-f(n)=f'(cn).
Then, f(n)-f(0) telescopes to f'(c1)+f'(c2)+....+f'(cn) ; cn in (n,n+1).
Now, f(n)-f(0) is bounded, and increasing, then f'(cn)>0 , so that the
partial sums f'(cj) are a monotone
(since f'(cn)>0 ) bounded sequence, which then converges ( to the LUB of the
sum) , so that the tail --and so the individual terms -- goes to zero.
> > > > 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.