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I'd love your help with this question:

Let $f$ be a continuous function in $[0,\infty)$, and $\lim_{x \to \infty }f(x)$ exist and finite.

I need to prove that $f(x)$ is bounded. Furthermore, if $\lim_{x \to \infty }f(x)=\lim_{x \to -\infty}f(x)$ and they are both finite, the function gets a maximum and minimum values.

What is the difference between being bounded and getting min and max? I know that under these conditions, this function is uniformly continuous.

Thank you

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@Nir: if this is homework, please tag it as such (if not, great - I don't assume that it is). The term "final" is unclear to me, are you translating it from some other language? Hebrew? If so, I might be able to help. –  Alon Amit May 9 '11 at 20:35
@Alon - how do you right קיים וסופי? And it is not homework, I need to refresh my memory in things I already studied, like this. –  user6163 May 9 '11 at 20:37
i think he means finite –  yoyo May 9 '11 at 20:40
@Nir: You are looking for "the limit exists and is finite". –  Asaf Karagila May 9 '11 at 20:40
@Nir: you also want to write $[0,\infty)$ for the domain rather than $[0,\infty]$. –  Alon Amit May 9 '11 at 20:54

3 Answers 3

up vote 1 down vote accepted

There is a difference between being bounded and achieving min or max. For example, $\frac{1}{x}$ is continuous and bounded on $[1,\infty)$, but doesn't achieve a min (clearly the infimum is 0).
As to your question: Use the definition of the limit at infinity to get $M$ such that for all $x>M$ we have $L-\varepsilon<f(x)<L+\varepsilon$. Then use what you know about functions continuous on $[0,M]$ to conclude your proof.
For the second part, use the same argument to say that the only interesting part is $[-M,M]$ for some $M>0$, and there the result follows.

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"Bounded" means that for some $M$ we have $|f(x)| \leq M$ for all $x \in \mathbb{R}$ (or $f(x) \leq M$ for "bounded above", or $f(x) \geq M$ for "bounded below"). Attaining a maximum means that the function actually has the value $M$ at some point $x_0$, with $M$ being the maximal value (so $f(x_0)=M$ for some $x_0$ while $f(x) \leq M$ for all $x$).

The function $f(x)=1/x^2$ is bounded below but never attains a minimum. Make sure you see that.

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$f$ is uniformly continuous (hence bounded) on say $[-N,N]$ so if $f$ were unbounded, there must be a sequence $x_n\to\pm\infty$ such that $f(x_n)\to\pm\infty$. this contradicts the existence of $\lim_{x\to\pm\infty}f(x)$.

the second part doesnt seem true: $1/(1+x^2)$ goes to zero at $\pm\infty$, but doesnt attain its minimum (of zero).

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You mean the existence of a finite limit at $f(\pm\infty)$. It is possible that there is no limit at all. –  Asaf Karagila May 9 '11 at 20:41

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