# Find the limit of the function with the given properties

• Nov 22nd 2011, 07:45 PM
My Little Pony
Find the limit of the function with the given properties
Let f: (0,1) --> R be any function with the following properties.

1) $\displaystyle \lim_{x\to 0}{f(x)} = 0$

2) $\displaystyle \lim_{x\to 0}({\frac{f(x)}{x} - \frac{f(x/2)}{x}}) = 0$

Then find: $\displaystyle \lim_{x\to 0}{\frac{f(x)}{x}$

I know that f(x) = sin(x^2) follows the above properties. Thus, $\displaystyle \lim_{x\to 0}{\frac{f(x)}{x} = 0$. But I can't prove this for any general function.

Any hintss to get me started?
• Nov 23rd 2011, 12:19 AM
chisigma
Re: Find the limit of the function with the given properties
Quote:

Originally Posted by My Little Pony
Let f: (0,1) --> R be any function with the following properties.

1) $\displaystyle \lim_{x\to 0}{f(x)} = 0$

2) $\displaystyle \lim_{x\to 0}({\frac{f(x)}{x} - \frac{f(x/2)}{x}}) = 0$

Then find: $\displaystyle \lim_{x\to 0}{\frac{f(x)}{x}$

I know that f(x) = sin(x^2) follows the above properties. Thus, $\displaystyle \lim_{x\to 0}{\frac{f(x)}{x} = 0$. But I can't prove this for any general function.

Any hintss to get me started?

Applying l'Hopital rule to condition 2) You obtain that $\displaystyle \lim _{x \rightarrow 0} f^{'} (x)=0$ so that is $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{x} = \lim _{x \rightarrow 0} f^{'} (x)=0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 23rd 2011, 12:32 AM
FernandoRevilla
Re: Find the limit of the function with the given properties
Quote:

Originally Posted by chisigma
Applying l'Hopital rule to condition 2)

We don't know a priori if $\displaystyle f$ is differentiable.
• Nov 23rd 2011, 11:25 AM
chisigma
Re: Find the limit of the function with the given properties
An alternative that doesn't require the derivative of f(x) starts setting $\displaystyle \lim_{x \rightarrow 0} \frac{f(x)}{x}= \lambda$. Now setting $\displaystyle \frac{x}{2}= \xi$ we have $\displaystyle \lim_{\xi \rightarrow 0} \frac{f(\xi)}{2 \xi}= \frac{\lambda}{2}$ and the condition 2) becomes...

$\displaystyle \lambda - \frac{\lambda}{2}=0$ (1)

... and the only value of $\displaystyle \lambda$ that satisfies (1) is $\displaystyle \lambda=0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$