Prove or disprove: $\lim_{x→0} f(x) = 0$

Question

Let $f : \mathbb{R} → \mathbb{R}$ be a function such that for any $r ∈ \mathbb{R}$ , we have

$$\lim_{n\rightarrow \infty }f\left ( \frac{r}{n} \right )= 0$$

Prove or disprove: $\lim_{x→0} f(x) = 0$

My claim: The statement is true because $r ∈ \mathbb{R}$.

In other words, I think since we can pick any $r ∈ \mathbb{R}$, with $\lim_{n\rightarrow \infty} \frac{r}{n}$, we can cover any neighbor of $0$. (Either rational or irrational numbers near $0$).

But, since the reasoning does not seem strong enough for me, I also thought of any counterexample of $f$. So, I tried to construct a discontinuous function at $x=0$, but I couldn't think of any counterexample.

Is the statement really true? Then, how should I change (or improve) my reasoning to write it in clear mathematical terms?

If the statement is false, what should I notice to have a counterexample?

Answer 1

The statement is false. Fix some $a\in (0,1)$ and define the (geometric) null sequence $(x_n)=(a^{n})=(a,a^2,a^3,...)$. Then define $f$ to be the indicator function on the set $\{x_n: n\in \mathbb{N}\}$, i.e., $f\colon \mathbb{R}\to \mathbb{R}$, $f(x)=1$ if $x=x_n$ for some $n\in \mathbb{N}$ and $f(x)=0$ otherwise. Then for any given $r\in \mathbb{R}$ we have $\frac{r}{n}=x_n=a^n$ for only finitely many $n$ (otherwise there was a subsequence $(n_k)$ such that $r=n_k a^{n_k}\to 0$ as $k\to \infty$, which yields $r=0$ contradicting the equation $\frac{r}{n}=a^n$). Therefore $f(\frac{r}{n})=0$ for all but finitely $n$, thus $\lim_{n\to \infty} f(\frac{r}{n})=0$. But clearly, $\lim_{n\to \infty}f(x_n)=1$. So, the limit $\lim_{x\to 0} f(x)$ does not exist.