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Does there exist a continuous and bounded on $\mathbb{R}$ function $f(x)$ such that $\lim\limits_{x\to +\infty} f(x)$ does not exist but there exists the limit of the sequence $\lim\limits_{n\to \infty} f(n)$ $\quad$ $(n\in \mathbb{N})$ ?

$\sin x$ does not work, but may be there is a similar function?

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    $\begingroup$ $f(x)=\sin(\pi x)$. $\endgroup$ – Omran Kouba Nov 4 '15 at 18:41
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Check $x \mapsto \sin(\pi x)$.

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It's not continuous, but: $$f(x)=\begin{cases}1,&x\in\Bbb N\\0,&x,\notin\Bbb N\end{cases}$$ works.

For continuous examples, $\sin(\pi x)$ works. (Note that $\sin(\pi n)=0$ for $n\in\Bbb N$.)

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