I am asked to determine whether the following set is bounded $$\{x \in \Bbb R \mid x^7 + 3e^x + \ln|x| < 0\}$$

I am asked to determine whether this set is bounded. I've checked the test answers after not being able to solve the problem, and it turns out it is.

I've been banging my head against a wall for the past 20 minutes with no luck; I can't find a way to determine whether this set is bounded, has only an upper/lower bound, has maxima or minima.

Since I'm having troubles solving similar problems as well, I'd really appreciate it if someone pointed out to me a consistent way to determine bonds, sup/inf, and maxima/minima of sets and functions, plus how to do it in this specific case.

Thank you!


Hint. Evaluate the limit $$\lim_{x\to -\infty}\left( x^7 + 3e^x + \ln|x|\right)= \lim_{x\to -\infty}x^7 \left( 1+ \underbrace{\frac{3e^x}{x^7}}_{\to 0} + \underbrace{\frac{\ln|x|}{x^7}}_{\to 0}\right).$$

  • $\begingroup$ That limit should equal ${-\infty}({-\infty}) = {+\infty}$ and since the limit for ${x\to +\infty}$ is ${+\infty}$ as well, there have to be two values, one negative and one positive, that don't meet the $<0$ condition, therefore the set must be bounded. Is this reasoning correct? $\endgroup$ – Samuele B. Oct 20 '19 at 14:22
  • $\begingroup$ The limit is ${-\infty}(1+0+0) = {-\infty}$ $\endgroup$ – Robert Z Oct 20 '19 at 14:23
  • $\begingroup$ yeah, my bad... So that means that the set has an upper bound but now a lower one? $\endgroup$ – Samuele B. Oct 20 '19 at 14:24
  • $\begingroup$ thank you. Just to make sure, $\frac{\ln |x|}{x^7} = 0$ when ${x \to -\infty}$ because $x^7$ is "faster" to get to minus infinity than the logarithm, correct? $\endgroup$ – Samuele B. Oct 20 '19 at 14:28
  • 1
    $\begingroup$ Yes, you are right again. $\endgroup$ – Robert Z Oct 20 '19 at 14:31

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