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Suppose $f$ is continuous. For all $x>0$, the limit of $f(nx)$ when $n$ goes to infinity is $0$.

Then please prove that the limit of $f(x)$ as $x$ goes to infinity is $0$. (I totally stuck at it) I think it suffices to show that f is uniformly continuous on $[0, \infty)$.

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    $\begingroup$ So for $x =1 $ you have $\lim_{n\to \infty} f(n) = 0$? $\endgroup$
    – Thomas
    Mar 21 '13 at 15:45
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Let $\epsilon >0$. Let $B_n$ be the set of positive reals numbers $x$ such that $\mid f(mx)\mid\leq\epsilon$ for all $m\geq n$. The hypothesis says that $(0,\infty)=\cup B_n$. The Baire category theorem implies that there exists $n$ such that $B_n$ contains an interval, say $(a,b)\subset B_n$. Then $\mid f(x)\mid\leq \epsilon$ for all $x\in \cup_{n\geq m} (na,nb)$. It is an easy exercise to prove that $ \cup_{n\geq m} (na,nb)$ contains $(N,\infty)$ for some large $N$. Then $\mid f(x)\mid\leq \epsilon$ for all $x\geq N$.

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    $\begingroup$ +1, but it seems you used quite a strong artillery with Baire. Let's see whether there will be other answers $\endgroup$
    – Ilya
    Mar 21 '13 at 16:26
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Since the real numbers are closed under multiplication, it suffices to show that there is some $n$ for which $x_1 = nx_2$, therefore $x_1$ can be represented as $nx_2$.

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  • $\begingroup$ Can you please say me, what to do next to conclude the proof? $\endgroup$
    – Tomás
    Mar 21 '13 at 17:29

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