# Application of MVT

1. Jan 15, 2008

### sinClair

1. The problem statement, all variables and given/known data
Let f be diff. on (0,infinity) If the limit of f'(x) as x->infinity and limit of f(n) as n->infinity both exist and are finite, prove limit of f'(x) as x->infinity is 0.
2. Relevant equations
Mean Value Theorem (applied below)
3. The attempt at a solution
Suppose a>0 and b>0. Then by mvt there exists c in (a,b) such that f'(c)=(f(b)-f(a))/(b-a).

Now taking the limit of both sides with respect to b as b->infinity, f'(c)=0 since the limit of f(n) as n->infinity is finite. Now, take the limit of both sides with respect to c as c->infinity and we have what we want?

Not sure if this does it or is clear because the presence of f(a) might turn limit into indeterminate form? But f(a), and a is finite so taking the limit of both sides still yields what we want. This seemed a little too "convenient"....

Thank you for looking.

2. Jan 15, 2008

### Vid

a isn't changing at all so f(a) is just some constant.

3. Jan 15, 2008

### sinClair

Yeah, so it's ok to take the limit of both sides twice with respect to different variables without something going wrong? So it's the limit of a limit or I guess taking one variable to infinity then another one at a time. Sorry this probably just sounds inane.

Last edited: Jan 15, 2008
4. Jan 15, 2008

### Vid

Well, instead of using the interval (a,b), try a neighborhood of b.

f'(c) = (f(b-e) - f(b+e))/2e

Take the limit as b goes to infinity. Since both f(b-e) and f(b+e) will go to the same limit, f'(c) = 0.

5. Jan 16, 2008

### sinClair

I think I see the problem: MVT itself shows that f'(c)=0 for some fixed c>0. But the problem asks to show that lim f'(x) as x-> infinity is 0.

Last edited: Jan 16, 2008