Fermat, Rolle and Lagrange

Prove: if $\displaystyle f(x)$ is differentiable on $\displaystyle (a,\infty)$ and $\displaystyle \lim_{x\to \infty} f'(x)=0$, then $\displaystyle \lim_{x\to \infty} \frac {1}{x}[f(2x)-f(x)]=0$.

How do I approach such an exercise? Any ideas would be appreciated.

Re: Fermat, Rolle and Lagrange

You know that for any $\displaystyle \epsilon > 0$ you can find an $\displaystyle x$ so that $\displaystyle x>N \Rightarrow |f'(x)| < \epsilon$. In particular, this says that for any $\displaystyle \epsilon > 0$ you can find an x so that $\displaystyle |f'(2x)-f(x)| < \epsilon x$. Can you see why that is? Can you then conclude anything about your limit?

Re: Fermat, Rolle and Lagrange

Quote:

Originally Posted by

**thesmurfmaster** In particular, this says that for any $\displaystyle \epsilon > 0$ you can find an x so that $\displaystyle |f'(2x)-f(x)| < \epsilon x$. Can you see why that is?

No I can't see that. Can you please explain why?

Re: Fermat, Rolle and Lagrange

Let's say that $\displaystyle |f'(x)| < \epsilon$ for all $\displaystyle x>N$ (we know by definition that we can find such an N for each $\displaystyle \epsilon$). From $\displaystyle f(x)$ to $\displaystyle f(2x)$ we're changing the input by x (2x-x=x). Since the absolute value of the derivative is always less than epsilon in this interval, we can definately not move further from f(x) than $\displaystyle x * \epsilon$ when we move x units ahead. Therefore $\displaystyle |f(x) - f(2x)| < x*\epsilon$.

Re: Fermat, Rolle and Lagrange

Let's say that $\displaystyle |f'(x)| < \epsilon$ for all $\displaystyle x>N$ (we know by definition that we can find such an N for each $\displaystyle \epsilon$). From $\displaystyle f(x)$ to $\displaystyle f(2x)$ we're changing the input by x (2x-x=x). Since the absolute value of the derivative is always less than epsilon in this interval, we can definately not move further from f(x) than $\displaystyle x * \epsilon$ when we move x units ahead. Therefore $\displaystyle |f(x) - f(2x)| < x*\epsilon$.

Re: Fermat, Rolle and Lagrange

Quote:

Originally Posted by

**thesmurfmaster** Since the absolute value of the derivative is always less than epsilon in this interval, we can definately not move further from f(x) than $\displaystyle x * \epsilon$ when we move x units ahead.

Why? How does the former lead to the latter? I'm sorry if this looks trivial for anyone, but it simply doesn't look so for me... Can you perhaps explain this in a different matter? Thanks again :)