# Differentiability and convergence.

1. Jan 10, 2009

### math8

Let f:(a,infinity)-->R (reals) be differentiable and let A and B be real numbers. Prove that if f(x)-->A and f '(x)-->B as x --> infinity, then B=0.

I would guess that the mean value theorem may be needed but I am not sure how to use it considering that we're dealing with x --> infinity.

2. Jan 10, 2009

### Dick

There's no such number as 'infinity' to put into the mean value theorem, sure. But if f(x)->A then clearly f'(x) must become very small as x becomes large, right? Write down the definition of f(x)->A, f'(x)->B. It means there is some number N such that for all x>N etc etc. Pick an interval in the range x>N and apply the MVT. Just TRY it.

Last edited: Jan 10, 2009
3. Jan 10, 2009

### math8

by drawing, I do understand what's going on, but writing the proof I get confused, this is what I've got so far:
f(x)-->A means for e>0 given, there is some number N such that for all x>N, we have |f(x)-A|<e
f '(x)-->B means there is some number M such that for all x>M, we have |f '(x)-B|<e

Now I let (x1,x2) C (N, inf.), f diff, and continuous, MVT applies and we get some z in (x1,x2) such that f(x2)-f(x1)/(x2-x1)=f '(z)
Now z > N implies that |f(z)-A|< e.

Now I don't see how to proceed :(

4. Jan 10, 2009

### Dick

Ok. Pick an N big enough that both |f(x)-A|<e and |f'(x)-B|<e for x>N. Let's just look at the interval [N,N+1]. |f(N+1)-f(N)|<2e, right? MVT says there is a c in [N,N+1] such that f'(c)=f(N+1)-f(N). But |f'(c)-B|<e. Do you see where this is going?

5. Jan 10, 2009

### math8

oh yeah, |f'(c)-B|<e, so -e<f'(c)-B<e but f'(c)=f(N+1)-f(N)<2e, hence B<f'(c)+e<3e.
Thus, since e was arbitrary, B=0.

6. Jan 10, 2009

### Dick

Right. |B|<3e. All you have to do is fix on some definite interval to apply the MVT.

7. Jan 10, 2009

### math8

right, thanks a lot.