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Prove or disprove: if $f$ is uniformly continuous on $[1,\infty)$ then there exists a limit, $\lim_\limits{n\to \infty}{f(x)}=L$ where $-\infty \le L\le \infty$.

I tried to find a counterexample but it would work. I would really appreciate your help.

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    $\begingroup$ $f(x) = \sin x$? $\endgroup$ – Ulrik Mar 4 '15 at 8:30
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Hint: $\sin \colon [1,\infty) \to \mathbb R$ is Lipschitz and hence uniformly continuous.

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