```Date: May 26, 2013 1:28 AM
Author: Bacle H
Subject: Re: Does this imply that lim x --> oo f'(x) = 0?

On Saturday, May 25, 2013 10:25:30 PM UTC-7, bacle...@gmail.com wrote:> On Friday, May 24, 2013 8:28:07 PM UTC-7, Graham Cooper wrote:> > > On May 25, 12:50 pm, William Elliot <ma...@panix.com> wrote:> > > > > > > On Fri, 24 May 2013, baclesb...@gmail.com wrote:> > > > > > > > On Friday, May 24, 2013 3:28:09 AM UTC-4, William Elliot 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.> > > > > > > >  How about this: with the same lay out as before: f(n+1)-f(n)=f'(cn).> > > > > > >> > > > > > > Give it up, counter examples have been presented.> > > >  Maybe you should check your counterexamples more carefully.Morover, you told me you did not see where monotonicity was used, andI showed you where, or that I produce an argument where monotonicitywas used, and this is what I did. So don't ask me for something and thencomplain when I answer your question.> > > > > > > > > > > > I think this one works..> > > > > > > > > > > > -1/(5+sin(x))/x/x> > > > > > > > > > > > > > > > > > http://www.wolframalpha.com/input/?i=-1%2F%285%2Bsin%28x%29%29%2Fx%2Fx> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Herc> > > > > > --> > > > > > www.BLoCKPROLOG.com
```