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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?

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    $\begingroup$ The answer would certainly be yes if $\lim_{n\to \infty} f(r/n) = 0$ was uniform on $r$. $\endgroup$ May 8, 2021 at 4:15
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    $\begingroup$ That is, if the starting $N$ of the definition of the limit does did not depend on $r$ $\endgroup$ May 8, 2021 at 4:17

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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.

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  • $\begingroup$ Can you please give me a little more details about the reason why we do have only finitely many n? I would like to know why we are considering $(n_k)$ to get a contradiction. $\endgroup$
    – john
    May 8, 2021 at 5:32
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    $\begingroup$ It is just reasoning by contradiction. If the equation held for infinitely many $n$, then you can construct a subsequence $(n_k)$ such that the eqation holds for each $n_k$ and $n_k\to \infty$ for $k\to \infty$. If this reasoning is not clear to you, you could even argue more directly: From $\frac{r}{n+1}=a^{n+1}$ and $\frac{r}{n}=a^n$, we get (by dividing the equations) $1-\frac{1}{n+1}=\frac{n}{n+1}=a$, and you can easily see that this equation can be true for only one $n$. Thus, in fact, the equation can only hold for at most two $n$. $\endgroup$
    – ym94
    May 8, 2021 at 11:20

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