# Fermat, Rolle and Lagrange

Printable View

• Jul 12th 2012, 03:27 AM
loui1410
Fermat, Rolle and Lagrange
Prove: if $f(x)$ is differentiable on $(a,\infty)$ and $\lim_{x\to \infty} f'(x)=0$, then $\lim_{x\to \infty} \frac {1}{x}[f(2x)-f(x)]=0$.

How do I approach such an exercise? Any ideas would be appreciated.
• Jul 12th 2012, 05:08 AM
thesmurfmaster
Re: Fermat, Rolle and Lagrange
You know that for any $\epsilon > 0$ you can find an $x$ so that $x>N \Rightarrow |f'(x)| < \epsilon$. In particular, this says that for any $\epsilon > 0$ you can find an x so that $|f'(2x)-f(x)| < \epsilon x$. Can you see why that is? Can you then conclude anything about your limit?
• Jul 12th 2012, 05:24 AM
loui1410
Re: Fermat, Rolle and Lagrange
Quote:

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

No I can't see that. Can you please explain why?
• Jul 12th 2012, 10:21 AM
thesmurfmaster
Re: Fermat, Rolle and Lagrange
Let's say that $|f'(x)| < \epsilon$ for all $x>N$ (we know by definition that we can find such an N for each $\epsilon$). From $f(x)$ to $f(2x)$ 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 $x * \epsilon$ when we move x units ahead. Therefore $|f(x) - f(2x)| < x*\epsilon$.
• Jul 12th 2012, 10:29 AM
thesmurfmaster
Re: Fermat, Rolle and Lagrange
Let's say that $|f'(x)| < \epsilon$ for all $x>N$ (we know by definition that we can find such an N for each $\epsilon$). From $f(x)$ to $f(2x)$ 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 $x * \epsilon$ when we move x units ahead. Therefore $|f(x) - f(2x)| < x*\epsilon$.
• Jul 12th 2012, 10:44 AM
loui1410
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 $x * \epsilon$ 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 :)