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Let $f$ be a differentiable a function in $\mathbb{R}$, and let $\lim _{x\rightarrow \infty}{f'(x)}=0$

Does $\lim_{x \rightarrow \infty}{f(x)}$ exist in the broad sense?

I'm really lost here. This exercise is from a section on MVT, and intuitively it seems to be correct, but I can't seem to find a lead. If someone could just give me a hint that would be great.

So far my best shot has been using Heine's definition of the limit, but no dice.

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  • $\begingroup$ Do you want to know when this limit would exist ?or do you want to know when it won’t ? $f(x)=\log(x)$ and $f’(x)=\frac{1}{x}$ is a case where limit won’t exist for $f(x)$ $\endgroup$
    – SagarM
    Jan 21, 2021 at 17:32
  • $\begingroup$ I want to know if it always exists in the broad sense, so those examples wouldn't work $\endgroup$
    – ShiNyyf
    Jan 21, 2021 at 17:35
  • $\begingroup$ Is there a formal definition of “broad sense”? $\endgroup$
    – SagarM
    Jan 21, 2021 at 17:38
  • $\begingroup$ I was sure that was the correct term but I can't find any evidence of that.now haha. Sorry, I just mean the limit can be + or - infinity aswell as finite $\endgroup$
    – ShiNyyf
    Jan 21, 2021 at 17:39
  • $\begingroup$ At best you can use L'Hospital's Rule and say that $f(x) /x\to 0$ as $x\to\infty $. But limit of $f$ can be anything including $\pm\infty $ and oscillation. $\endgroup$
    – Paramanand Singh ♦
    Jan 22, 2021 at 7:27

1 Answer 1

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Intuitively, what you want here is a function that iterates between (at least) two values and also gets straighter as time passes.

You can think of the usual sine function as a spring, and imagine stretching it to the right. Try to understand why this make the function get straight as $x$ increases.

Consequently, $\sin(\sqrt x)$ would work.

I suggest you draw it (using a graphing website) to see why.

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  • $\begingroup$ Brilliant! Thank you! That was my thought process, but I couldn't find a function that held under those conditions. Every function I tried ended up either converging or the derivative didn't go to zero, so after several tries I assumed it was true and tried proving it, glad I didn't stick with it. $\endgroup$
    – ShiNyyf
    Jan 21, 2021 at 17:36
  • $\begingroup$ @shinyff you should stop thinking that only combinations of arithmetic, trigonometric, and exponential operations are valid functions. It's trivial to know that a counterexample to your conjecture exists because you can just draw it (and you can make such arguments rigorous by defining piecewise linear functions and smoothing them by convolutions). In this case this happens to be not necessary because there is an easy analytic counterexample, as this answer shows, but that's just a coincidence $\endgroup$
    – Bananach
    Jan 21, 2021 at 17:43
  • $\begingroup$ You're absolutely right, intuitively it actually made sense to me that the statement would be right, and when I couldn't seem to find a counterexample I went to proving. This question is from a 100 level course so usually counterexamples are either elementary functions, or piecewise functions made of elementary function compositions, as at this point we don't really know a lot more than that. $\endgroup$
    – ShiNyyf
    Jan 21, 2021 at 17:53

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