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?

  • $\begingroup$ 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
  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.
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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