Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a continuous function $f\colon \mathbb R \to \mathbb R$ and the fact that $ \lim_{x\rightarrow \infty} f(x)$ and $ \lim_{x\rightarrow -\infty}f(x)$ exist (finite), prove that $f$ is bounded. I understand why it's true, but I have no idea how to formally prove this. I'd appreciate the help.

share|improve this question

2 Answers 2

There exists $x_1$ with $|f(x)-\lim_{y\to\infty}f(y)|<1$ for all $x>x_1$. There exists $x_2$ with $|f(x)-\lim_{y\to-\infty}f(y)|<1$ for all $x<x_2$. And $f$ is bounded on the compact interval $[x_2,x_1]$.

share|improve this answer
    
It's bounded because of the Bolzano–Weierstrass theorem (since it's a closed interval and continuous)? –  Shookie Jul 18 '13 at 14:17
    
The image of a compact by a continuous function is a compact (provided some very mild conditions on the topology); in this case (reals), it's the extreme value theorem, which can be proven using the Bolzano–Weierstrass theorem. –  Clement C. Jul 18 '13 at 14:27

Clearly, it is sufficient to prove that $f$ is bounded on $\mathbb{R}_+$, the problem being symmetric.

Let $\ell\stackrel{\mathrm{def}}{=} \lim_{+\infty} f$; wlog, $\ell=0$ (just by "considering" $g\stackrel{\mathrm{def}}{=}f-\ell$ instead).

  • By definition [1], there exists $A\geq 0$ such that $\forall x \geq A$ $|f(x)|\leq 1$.
  • Moreover, $f$ being continuous on $[0,A]$, it is also bounded there: let $M=\max_{x\in[0,A]} |f(x)|$.

Setting $$M^\prime\stackrel{\mathrm{def}}{=}\max(M,1)$$ we thus have that for all $x \geq 0$, $|f(x)| \leq M^\prime$; that is, $f$ is bounded on $\mathbb{R}_+$.

[1] definition of the limit, taking $\varepsilon=1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.