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Suppose $f: \mathbb{R} \to \mathbb{R}$. I must determine whether the following statements are true:

  1. If $f$ is continuous on $\mathbb{R}$ and not bounded then $\lim_{x\to \infty} f(x)$ is either $\infty$ or $-\infty$.

  2. If $f$ is strictly increasing and is not bounded from below then $\lim_{x\to -\infty} f(x)=-\infty$

I believe that both are true. In 1. I think that if the function is not bounded and has $\lim_{x\to \infty} f(x)=L$ then it must tend to $\pm$infinity at some point. But if $f$ tends to $\pm \infty$ at any point then it can't be continuous there. So it must tend to $\pm \infty$ at $\infty$. In 2. I can't find a counterexample but I have a hard time proving the statement. Intuitively, I think that it is true. Am I correct?

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  • $\begingroup$ Hint for the first one: think about $x \sin(x)$, in particular think about this function at $\pi/2,3\pi/2,5 \pi/2,\dots$. The second one is true; try to prove it. $\endgroup$ – Ian May 18 '15 at 18:32
  • $\begingroup$ @user241601: The term "monotonic" appears in the title but not in the question body. Are you assuming $f$ is monotonic in 1.? (Incidentally, your proposed proof for 1. isn't correct, even assuming $f$ is monotonic: If $f(x) \to L$ as $x \to \infty$, you could still have unboundedness as $x \to -\infty$.) $\endgroup$ – Andrew D. Hwang May 18 '15 at 18:59
  • $\begingroup$ @user86418 - in proposition #2 $f$ is strictly increasing = monotonic. $\endgroup$ – user241601 May 18 '15 at 19:21
  • $\begingroup$ @user241601: Of course, but your question title suggests you're assuming $f$ is monotonic throughout. In Proposition 1, however, your proposed counterexample is not monotonic, which may mean it isn't a counterexample. My previous comment addressed Proposition 1, assuming $f$ is required to be monotonic. If that's not your hypothesis, please disregard. $\endgroup$ – Andrew D. Hwang May 18 '15 at 20:06
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  1. Consider $f(x)=x\sin x$.
  2. This is true. Hint for a proof: For any $L\in \mathbb R$, there exists $c\in \mathbb R$ such that $f(c) < L$, since $f$ is not bounded below. Also, if $f(c)< L$, then $f(x) < L$ for all $x\leq c$, since $f$ is strictly increasing.
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  • $\begingroup$ Ah, right. There are functions with infinitely increasing "oscillations" which don't have a limit. Thank you. $\endgroup$ – user241601 May 18 '15 at 18:35

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