# Differentiable function, limits, sequence

1. Feb 6, 2008

### alligatorman

f is differentiable on $$(a,\infty)$$ and

$$\lim_{x\to\infty}\frac{f(x)}{x}=A$$

I am trying to prove that there exists a sequence $$\{x_n\}, x_n\rightarrow \infty,$$ such that $$f'(x_n)\rightarrow A.$$

Any help would be appreciated.

2. Feb 6, 2008

### Hurkyl

Staff Emeritus
Can you give a heuristic explanation why it should be true?

3. Feb 7, 2008

### mathwonk

then what is the limit of [f(x) - f(a)]/[x-a], for any a?

4. Feb 7, 2008

### alligatorman

By MVT, it would equal f'(c), where c is in (a,x). But I don't see how I can get a sequence from this fact.

As x gets larger, there is always a c_n in (a,x) such that f'(c)=[f(x) - f(a)]/[x-a]. Perhaps it can be said that the sequence of x_n approaches A, but I don't know how to get there.