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We have the continuous function $f\colon [a;\infty) \to \mathbb{R}$ . How can i prove that it is uniformly continuous? The function has a finite limit in $+\infty$.

I know that for a pair of $a,b \in \mathbb{R}$, $f\colon[a;b) \to\mathbb{R}$ is continuous if the $ \lim\limits_{\substack {x\to b \\x

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    But that is false. $f\colon x\mapsto x^2$ is not uniformly continuous on, say, $[1,\infty)$. Did you forget an assumption? – 2017-02-04
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    Not every continuous function $[a,\infty)\to\Bbb R$ is uniformly continuous, consider for example the function $x\mapsto x^2$. Do you have any conditions? – 2017-02-04
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    Hmm i understand, thank you for your fast and prompt answers! – 2017-02-04
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    https://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem ... continuous functions on compact are uniform continuous. – 2017-02-04
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    Wouldn't Heine - Cantor be impossible to use as $[1;+\infty)$ is not a compact set? – 2017-02-04
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    If you know that $f$ has a finite limit at $\infty$, then things are a bit different. Did you have that in mind? (and to refer to @rtybase comment: this indeed does *not* apply to $[a,\infty)$.) – 2017-02-04
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    Not really,can you elaborate? – 2017-02-04
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    Yes! That was the assumption that i did not write! I forgot to mention that $f(x)$ has a finite limit in $\infty$ ! – 2017-02-04
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    If $\lim_\infty f = \ell \in\mathbb{R}$, then for every $\varepsilon>0$ you have a value $A$ such that $f(x)\in[\ell-\varepsilon,\ell+\varepsilon]$, and therefore $f(x)$ and $f(y)$ are very close whenever $x,y\in I=[A,\infty)$. Moreover, $[a,A]=J$ is a compact, so $f$ is uniformly continuous on $J$. Combining the two suitably, you'll get $f$ uniformly continuous. @Eduard6421 – 2017-02-04
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    Thank you! Could you post it again as an answer? I would like to mark the question as answered! – 2017-02-04
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    @Eduard6421 Yes I know $[a;+\infty)$ is not compact and you can't apply that theorem unless you have more conditions over $f$ ;) – 2017-02-04

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Following the exchange in the comments, posting what what there as answer.

  • This is false without further assumptions, as the example of $f\colon x\in[a,\infty)\mapsto x^2$ shows.

  • under the additional assumption that $\lim_\infty f$ exists (name it $\ell\in\mathbb{R}$), then this becomes a standard exercise (most likely asked several times on this website). The idea is to use

    • the fact that for $x,y$ "big", then $\lvert f(x) - f(y) \rvert \leq \lvert f(x)-\ell\rvert + \lvert \ell - f(y)\rvert$ and both terms of the RHS are very small since $f\to_\infty \ell$;
    • and for $x,y$ "small", well they are both in a compact interval of the form $[a,A]$, and then anyway $f$ is uniformly continuous on this compact by the Heine—Cantor theorem.

    More precisely, here is the outline:

    1. for every $\varepsilon>0$ you have a value $A_\varepsilon$ such that $f(x)\in[\ell-\frac{\varepsilon}{2},\ell+\frac{\varepsilon}{2}]$, and therefore by the above $f(x)$ and $f(y)$ satisfy $\lvert f(x) - f(y) \rvert \leq \varepsilon$ whenever $x,y\in I=[A_\varepsilon,\infty)$.

    2. But then, $[a,A_\varepsilon]=J$ is a compact, so $f$ is uniformly continuous on $J$, and there exists $\delta > 0$ such that for any $x,y\in J$ such that $\lvert x-y\rvert < \delta$ we have $\lvert f(x) - f(y) \rvert \leq \varepsilon$ as well.

    3. The last case to deal with is when $x\in I,y\in J$: then, use continuity at $A_\varepsilon$.

    Combining the three suitably, you'll prove (as $\varepsilon$ was arbitrary) that $f$ uniformly continuous on $[a,\infty)$.