# How to use mean vaule theorem here

1. Jan 18, 2009

### transgalactic

i got this question:
there is a function f(x) which is differentiable on (a,+infinity)
suppose lim [f(x)]/x =0 as x->+infinity
prove that lim inf |f'(x)|=0 as x->+infinity ?

does this expression lim inf f'(x)=0 has to be true
if not
present a disproving example

?

i was present whis this solution but i didnt quite understand it.
"First consider $\lim_{m\to\infty}\frac{f(2m)}{m}$, let $2m=x$ and this limit becomes $2\lim_{x\to\infty}\frac{f(x)}{x}=0$. So $\lim_{x\to\infty}\frac{f(2x)}{x}$ exists and equals $0$. So $\lim_{x\to\infty}\frac{f(x)}{x}=0\implies \lim_{x\to\infty}\frac{f(2x)-f(x)}{x}=\lim_{x\to\infty}\frac{f(2x)-f(x)}{2x-x}=0$. So now consider the interval $[x,2x]$ and apply the MVT letting x approach infinity."

mean value theorem says $f'(c)=\lim_{x\to\infty}\frac{f(2x)-f(x)}{2x-x}=0$
i dont know what the value of c??

and it doesnt prove
lim inf |f'(x)|=0 as x->+infinity

??

2. Jan 18, 2009

### Staff: Mentor

The theorem says that if f is continuous on [a, b], and differentiable on (a, b), then there exists some number c in (a, b) such that
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

The MVT doesn't tell you the value of c or how to find it; it just says that such a number exists.

I believe that what you're reading is saying that since f'(c) = 0 for some number c in [x, 2x], then this is also true in the limit as x approaches infinity.

3. Jan 18, 2009

### transgalactic

so how to prove that
lim inf |f'(x)|=0 as x->+infinity