Find the limit of the function with the given properties

**My Little Pony** · November 22nd 2011, 07:45 PM

Let f: (0,1) --> R be any function with the following properties.

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

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

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

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

Any hintss to get me started?

**chisigma** · November 23rd 2011, 12:19 AM

Originally Posted by My Little Pony

Let f: (0,1) --> R be any function with the following properties.

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

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

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

I know that f(x) = sin(x^2) follows the above properties. Thus, $\lim_{x\to 0}{\frac{f(x)}{x} = 0$ . 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 $\lim _{x \rightarrow 0} f^{'} (x)=0$ so that is $\lim _{x \rightarrow 0} \frac{f(x)}{x} = \lim _{x \rightarrow 0} f^{'} (x)=0$ ...

Kind regards

$\chi$ $\sigma$

**FernandoRevilla** · November 23rd 2011, 12:32 AM

Originally Posted by chisigma

Applying l'Hopital rule to condition 2)

We don't know a priori if $f$ is differentiable.

**chisigma** · November 23rd 2011, 11:25 AM

An alternative that doesn't require the derivative of f(x) starts setting $\lim_{x \rightarrow 0} \frac{f(x)}{x}= \lambda$ . Now setting $\frac{x}{2}= \xi$ we have $\lim_{\xi \rightarrow 0} \frac{f(\xi)}{2 \xi}= \frac{\lambda}{2}$ and the condition 2) becomes...

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

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

Kind regards

$\chi$ $\sigma$

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