If $f(x/n)\to0$ when $n\to\infty$, for every $x$ and $f$ is continuous, then $f(x)\to0$ when $x\to0$, or not? Let $f:(0,\infty)\to \Bbb R$ be continuous and such that for each $x&gt;0$ the sequence $\{f(\frac{x}{n})\}\to 0$.


  Does it imply that $\lim_{x\to 0^+} f(x)=0$?


My try:Fix $x&gt;0$.Since $f$ is continuous then $\lim _{n\to \infty}f(\frac{x}{n})=f(\lim_{n\to \infty} \frac{x}{n})=f(0)\implies f(0)=0$

Since $f$ is continuous and $f(0)=0\implies \lim _{x\to 0^+} f(x)=0$

Is the solution correct?In the book it is  given to use the Baire Category Theorem.
Expanding on Moya's comment, here is a big hint.

Fix $\epsilon&gt;0$. Let 
$$
K_n = \{x&gt;0:|f({x}/m)|\le\epsilon \text{ for all } m\ge n\}
$$
The idea is that for all $x$, the sequence $f(x/n)$ gets close 0 eventually, so we define $K_n$ to be the set of $x$ for which is this sequence is close to zero specifically at time $n$.

Next, prove that


The sets $K_n$ are closed.
The union of $K_n$ is $\mathbb R^+$.


Since (one form of) the Baire Category Theorem states that $\mathbb R^+$ cannot be written as a countable union of nowhere dense closed sets, you can conclude that one of the sets $K_n$ is somewhere dense, which for closed sets means it contains an interval $[x_1,x_2]$, for some $x_1&lt;x_2$. 

This is huge progress. Now, we have series of solid blocks of points $$[x_1/n,x_2/n],[x_1/(n+1),x_2/(n+1)],[x_1/(n+2),x_2/(n+2)]\dots,$$ where for any $x$ in these intervals, $|f(x)|&lt;\epsilon$. Use this fact to find a $\delta&gt;0$ such that $|f(x)|&lt;\epsilon$ whenever $|x|&lt;\delta$.
I'll give you an outline:

Start with the definition of a limit.

$$\forall \epsilon \text{ } \exists \delta \text{ s.t. } |f(x)|\leq \epsilon \text{ whenever } x\in (0,\delta).$$

And the definition of $\{f(\frac{x}{n})\}\to0$

$$\forall \epsilon \text{ } \exists N \text{ s.t. } |f(\frac{x}{n})|\leq \epsilon \text{ whenever } n&gt;N.$$

And continuity (*edit, accidentally wrote the definition for uniform continuity)

$$\forall \epsilon,x \text{ } \exists \delta \text{ s.t. } |f(x)-f(c)|\leq \epsilon \text{ whenever } |x-c|&lt;\delta.$$

First force $|f(x)|\leq \frac{\epsilon}{2}$, then force $|f(x)-f(c)|\leq \frac{\epsilon}{2}$, that gives that $f(c)&lt;\epsilon$.  You'll have to be a bit more rigorous with the deltas, let me know if I should elaborate.