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Assume $f:(0, \infty) \rightarrow \mathbb{R}$ is continuous. Also assume $\lim_{x \rightarrow 0}f(x)$ and $\lim_{x \rightarrow \infty} f(x)$ exist and are finite. Prove that $f(x)$ is bounded.

Proof:

Suppose $f(x)$ is not bounded on $(0,\infty)$. Since $\lim_{x \rightarrow 0}f(x)$ and $\lim_{x \rightarrow \infty} f(x)$ exist and are finite, we may assume further that there is some $x_0 \in \mathbb{R}$ such that $\lim_{x \rightarrow x_0}f(x)=\pm \infty$. But, since $f$ is continuous on $(0, \infty)$, it is continuous on $[x_0-1,x_0+1]$, which is compact. Then $f$ must be bounded on $[x_0-1,x_0+1]$, a contradiction. Thus no such $x_0$ exists.

Have I missed anything or does this cover everything? Thanks!

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  • $\begingroup$ Well, you have the right idea, but you need to justify that there is some $x_0$ such that $f$ tends to $\pm \infty$. You probably need to start with an appropriate compact set first. As an aside, it is probably easier to provide a 'constructive' proof. $\endgroup$
    – copper.hat
    Dec 5 '14 at 3:09
  • $\begingroup$ Thanks I forgot to put that $f$ is continuous. I edited the question. $\endgroup$
    – Sam
    Dec 5 '14 at 3:09
  • $\begingroup$ Got any ideas for that constructive proof? $\endgroup$
    – Sam
    Dec 5 '14 at 3:13
  • $\begingroup$ I added an approach below. $\endgroup$
    – copper.hat
    Dec 5 '14 at 3:16
  • $\begingroup$ Also look at Stone–Weierstrass theorem. $\endgroup$
    – Tacet
    Dec 5 '14 at 3:20
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Here is another approach: Let $L_0 = \lim_{x \downarrow 0} f(x), L_\infty = \lim_{x \to \infty} f(x)$. By definition of the limit we have some $\delta>0$ and $N>0$ such that if $x \in (0, \delta)$, then $|f(x)| \le 1+|L_0|$ and if $x > N$, then $|f(x)| \le 1 + |L_\infty|$.

Since $f([\delta,N])$ is compact it is bounded, hence there is some $M$ such that $|f(x)| \le M$ for $x \in [\delta,N]$.

Let $L = \max(1+|L_0|, 1 + |L_\infty|, M)$ to finish.

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Hint: Consider the map $g:(0,1)\to\mathbb R$ given by $g(x)=\tan\frac{\pi x}{2}$.

Show that $f\circ g$ has a continuous extension $h$ to $[0,1]$.

Then the image $h([0,1])$ is compact and therefore bounded.

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