$\lim\limits_{n\to\infty}f\left(\frac{x}{n}\right)=0$ for every $x > 0$. Prove $\lim\limits_{x \to 0}f(x)=0$ Function $f: (0, \infty) \to \mathbb{R}$ is continuous. For every positive $x$ we have $\lim\limits_{n\to\infty}f\left(\frac{x}{n}\right)=0$. Prove that $\lim\limits_{x \to 0}f(x)=0$. I have tried to deduce something from definition of continuity, but with no effect.
 A: The proof relies on the following lemma:
Let $I \subset (0,\infty)$ be a closed bounded interval. Let $U \subset (0,\infty)$ be an open subset accumulating at $0$. Then there exists some integer $N \geq 2$ and some closed interval $J \subset U$ such that $N \cdot J \subset I$. 
Sketch of proof: take some $x \in U$ with $x$ lower than the length of $I$ and the minimum of $I$ . For some integer $N \geq 2$, $Nx \in \overset{\circ}{I}$, thus, there exists some compact interval $J \subset U$ such that $N \cdot J \subset I$. 
Now, assume $f$ does not go to $0$ at $0$. Then, for some $\epsilon > 0$, $U=\{|f| > \epsilon\}$ is an open subset of $(0,\infty)$ accumulating at $0$.
By iterating the lemma, we can construct sequences of compact intervals $(I_n)$ and integers $(N_n)$ such that $N_{n+1}I_{n+1} \subset I_{n-1}$, $N_n \geq 2$.
Now, let $A_n=N_1 \ldots N_n$, then $A_n \rightarrow \infty$ and $K_n=A_n \cdot I_n$ is a non-increasing sequence of nonempty compact intervals in $I_0$. Thus, there exists $x$ that is in every $K_n$. 
So, $x/A_n \in I_n \subset U$, for all $n$, thus $|f(x/A_n)| > \epsilon$ for all $n$, a contradiction.
A: This is a standard application of Baire Category Theorem (BCT). Since $(0,\infty)=\cup_n \{x:|f(\frac x k)| \leq\epsilon \,\forall k \leq n\}$ there exists $n$ such that $\{x:|f(\frac x k)| \leq \epsilon \, \forall k \geq n\}$ contains some open interval $(a,b)$. [ Because $(0,\infty)$, being an open subset of $\mathbb R$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $f(\frac x n)$ to $0$ on some open interval from which the result follows easily. [ Details: let $0<\delta <b-a$ and $\delta < \frac a n$. Then, for any $x <\delta $ the interval $(\frac a x,\frac b x)$ has length exceeding $1$, so it contains an interger $k$. Hence $kx \in (a,b)$. Also, $k >\frac  a x > \frac a {\delta} > n$ so $|f(x)|=|f(\frac {kx} k)| <\epsilon$].
