$\lim_n{f(\frac{x}{n})}=0$, for all $x\in (0,1)$ implies $lim_{x\to0}f(x)=0$ (edit, $f$ is not continuous) Let be $f:(0,1)\rightarrow \mathbb{R}$. For all $x\in (0,1)$ I have that $\lim_n{f(\frac{x}{n})}=0$.
It's true that  $lim_{x\to0}f(x)=0$?
And if $f$ is continuous?
For the second point I just use the definition of continuity and limits:
Let be $x_0\in(0,1)$ and let be the succession $\{f(\frac{x_0}{n})\}$, and I know that for all $\varepsilon_1>0$ $\exists N>0$ such that if $n>N$ then $|f(\frac{x_0}{n})|<\varepsilon_1$.
$f$ is continuos in $\frac{x_0}{n}$, so for all $\varepsilon_2>0$ $\exists \delta_2>0$ such that if $|x-\frac{x_0}{n}|<\delta_2$ then $|f(x)-f(\frac{x_0}{n})|<\varepsilon_2$.
So I have that $|x|\le|x-\frac{x_0}{n}|+|\frac{x_0}{n}|<\delta_2+|\frac{x_0}{n}|=\delta^*$, because $x_0$ is fixed.
On the other hand $|f(x)|\le|f(x)-f(\frac{x_0}{n})|+|f(\frac{x_0}{n})|<\varepsilon_1+\varepsilon_2$.
But I don't know how solve the second. I try to prove that is false using many non-continuous function. but I failed.
(Sorry for my bad English, I'm Italian) 

I will provide a proof for the continuous case. This result requires Baire Category Theorem. If $\epsilon >0$ then $(0,1)=\cup_n A_n$ where $A_k=\{x:|f(\frac x n )| \leq \epsilon \forall k \geq n\}$. Since $(0,1)$ is of second category it follows that there is some interval $(a,b)$ and some $n_0$ such that $|f(\frac x n )| <\epsilon$ whenever $a<x<b$ and $ n \geq n_0$ $\cdots (1)$. Now the length of the interval $(\frac a x, \frac b x)$ is greater than $1$ whenever $0<x<b-a$. Hence there is an integer $n$ in this interval. This integer is also greater than $n_0$ if $\frac a x >n_0$ or $x <\frac a {n_0}$. Replacing $x$ by $nx$ in (1) we get $|f(x)| <\epsilon$ whenever $x$ is sufficiently small.
