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Let $f: \Bbb R \to \Bbb R$ a continuous, non-constant and periodic function, such that $T \in \Bbb R-\Bbb Q$ is one of its periods. Show that the sequence $f(n)$ cannot be convergent, where $n=0,1,2, \dots$

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    Hint: Since $f$ is periodic, for each $n$ there exists $t \in [0,T[$ with $f(t) = f(n)$. The set of such $t$'s are dense in $[0,T[$. Hence if $f(n)$ converges, $f$ is constant. – 2017-01-14
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    Can you be more specific, please? – 2017-01-14
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    I remember that the user Hagen Von Eitzen gave an answer to this question yesterday. Did you take a look at it? – 2017-01-14
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    It is too hard for my level of knowledge in this subject. – 2017-01-14
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    @HenryW. Can you help me please? – 2017-01-15

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$$\limsup_{n\to\infty}f(n)=\sup f$$

$$\liminf_{n\to\infty}f(n)=\inf f$$

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    Why is this true? – 2017-01-14