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Let $f:[0,\infty) \rightarrow \mathbb R$ be a function such that for any positive $a$ the sequence $f(an)$ converges to zero.Does the limit $lim_{x\rightarrow \infty} f(x)$ exist?

My problem is should I look for a counter example or try to prove it in which I have not succeeded till now.

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  • $\begingroup$ Generally, if you are unsure if something is true, you should spend half your time trying to prove it, and the other half trying to disprove it. Eventually you will succeed. $\endgroup$ – Mike Earnest Oct 25 '14 at 3:02
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    $\begingroup$ Did you forget to include the condition that $f$ is continuous? If this is not a requirement, then the statement is false (define $f(\sqrt{p})=1$ for any prime number $p$, $f(x)=0$ otherwise). $\endgroup$ – Mike Earnest Oct 25 '14 at 3:36
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When you add the condition that $f$ also be continuous, it turns out to that $f(x)\to0$. Here is a sketch of the proof.

Suppose $\lim_{x\to\infty}f(x)\neq 0$. Then there will be some $\epsilon>0$ and a sequence of points $x_k\to\infty$ for which $f(x_k)>2\epsilon$. Since $f$ is continuous, there is a closed interval $I_k$ around each point where $f(x)>\epsilon$ for $x\in I_k$.

For any $E\subset \mathbb{R}$, $c\in\mathbb{R}$, let $c\cdot E$ denote $\{cx\,|\,x\in E\}$.

Let $E_0=[0,1]$. Define a sequence of closed intervals $E_n$ for $n\ge1$ recursively as follows: choose some integer $m$ large enough so that, for some $k$, $(m\cdot E_{n-1})\cap I_k$ is a nontrivial closed interval (not just a point). You can show such an $m$ exists. Then, for that $k$, define $E_n=\frac1m\cdot( (m\cdot E_{n-1})\cap I_k)$. You may have to also insist if you use $I_k$ you choose for $E_n$ was not used for any previous $E_m$.

Since the $E_n$ are nested closed intervals, their intersection is nonempty. You can then show that, for any point $e$ in their intersection, the sequence $f(e),f(2e),f(3e)\dots$ will not approach zero, since the sequence $e,2e,3e\dots$ will hit infinitely many intervals $I_k$.

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I don't believe this is true.

Let $\{x_n\}$ be a sequence of distinct real numbers greater than 1 which are not multiples of each other and let $f(x_i*n) = 1$ if $i \le n$ and $f(x_i*n) = 0$ if $i \gt n$. Now suppose there was an $L$ such that $|f(x)| < \epsilon$ if $x > L$, this is not possible because for any $L$ we would have some $x_i*n \gt L$ where $f(x) = 1$.

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  • $\begingroup$ err, is this even continuous? $\endgroup$ – quasicompactscheme Oct 25 '14 at 3:46
  • $\begingroup$ Does it have to be? $\endgroup$ – integrator Oct 25 '14 at 4:01
  • $\begingroup$ for some reason I thought it had to be, I guess not $\endgroup$ – quasicompactscheme Oct 25 '14 at 12:36

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