# Limit of Function that Satisfies Simple Property [duplicate]

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Limit of a continuous function

Let $f$ be a function that is continuous and real on $[0, \infty]$ such that $\lim_{n \to \infty} f(na) =0$ for all $a>0$. What can be said about $\lim_{x \to +\infty} f(x)$?

Now I was told the answer is $0$ for every $f$ that satisfies the condition. But I do not know how to prove this. Can anyone help me please? Thank you!

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## marked as duplicate by Noah Snyder, Thomas, Cameron Buie, Davide Giraudo, DidDec 9 '12 at 16:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

That is a nice application of the Baire category theorem. For $\epsilon > 0$ let $A_n = \bigcap_{m\ge n} \left\{x \in \mathbb R \biggm||f(mx)| \le \epsilon \right\}$ Now $A_n$ is closed as intersection of closed sets ($f$ is continuous), and by assumption on $f$, the union $\bigcup_n A_n = (0,\infty)$. Now by Baire, one of the $A_n$ has an interior point, so there are $N \in \mathbb N$, $\delta > 0$ and $a \in (0,\infty)$ with $(a-\delta, a+\delta) \subseteq A_N$.

$\def\abs#1{\left|#1\right|}$ Let $m \ge N$ with $m \delta > a$, for any $s > ma$, choose $k \in \mathbb N$ such that $\abs{s-ka}$ is minimal (note that $m \le k$), we have $\abs{s - ka} < a < m\delta \le k\delta$ therefore $s \in k(a - \delta, a + \delta)$ and $|f(s)| \le \epsilon$.

So $f(x) \to 0$, $x\to\infty$.

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+1 Very nice answer... – copper.hat Nov 3 '12 at 23:20