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.


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 $[x_2,x_1]$.

  • $\begingroup$ It's bounded because of the Bolzano–Weierstrass theorem (since it's a closed interval and continuous)? $\endgroup$
    – matanc1
    Jul 18 '13 at 14:17
  • $\begingroup$ 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. $\endgroup$
    – 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 $\varepsilon=1$.


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