Suppose we have a differentiable real-valued function $f(x)$. The task is to prove, that if $\lim_{x\to\infty}f'(x) = 0$ then $\lim_{x\to\infty}f(x)/x = 0$, and, conversely, if $\lim_{x\to\infty}f(x)/x = 0$ than if the $\lim_{x\to\infty}f'(x)$ exists, than it's equal to zero.

I've tried using Lagrange theorem several times, but it didn't help. Could you please suggest a proof, or maybe help me to show that this fact is not true (though it seems to be true)?

  • $\begingroup$ You could use de l'hopital $\endgroup$ – Nathanael Skrepek May 6 '17 at 8:35
  • $\begingroup$ To use it shouldn't I get the $0/0$ or $infty/infty$? $\endgroup$ – Shot May 6 '17 at 8:37
  • $\begingroup$ You are right! My fault $\endgroup$ – Nathanael Skrepek May 6 '17 at 8:40
  • 1
    $\begingroup$ @Shot You can use l’Hôpital also in the form $\text{whatever}/\infty$. $\endgroup$ – egreg May 6 '17 at 9:32

They are not equivalent. Take $f(x)=\sin(x)$, then $$\lim_{x\to+\infty} \frac{\sin(x)}{x}=0$$ but $\lim_{x\to+\infty} \cos(x)$ does not exist.

However if $\lim_{x\to\infty}f'(x)$ exists then they are equivalent.

If $\lim_{x\to\infty}f(x)/x = 0$ and $\lim_{x\to\infty}f'(x)$ exists then it is zero by Lagrange theorem: there is $n<t_n<2n$ such that $$2\frac{f(2n)}{2n}-\frac{f(n)}{n}=\frac{f(2n)-f(n)}{2n-n}=f'(t_n)$$ By taking the limit as $n$ goes to $\infty$, the LHS goes to zero which implies that $f'(t_n)\to 0$. Hence $\lim_{x\to\infty}f'(x)$ has to be zero.

As regards the other implication, if $f$ is bounded then the result is trivial otherwise we apply Hopital rule .

  • $\begingroup$ Thank you, but I still have 2 remarks: 1) I've stated that $lim f'(x)$ should exist, but in your case it's not. 2) Why could I use l'Hopital? $\endgroup$ – Shot May 6 '17 at 8:38
  • $\begingroup$ If the $lim f(x)$ is not infinity, there are two cases, it could exist and be finite (and this is really trivial), or it could not exist at all, and still f(x) could be unbounded. Or couldn't it? And still no clue why could we use the l'Hopital rule, since there could be no undefinite fraction. $\endgroup$ – Shot May 6 '17 at 8:48
  • $\begingroup$ Is it clear now? if $f(x)$ is bounded then $|f(x)|/x\leq M/x\to 0$. $\endgroup$ – Robert Z May 6 '17 at 8:54
  • $\begingroup$ Thank you, very precise answer, but could you please clarify one more thing: you give an interesting form of Lagrange theorem, I know only the following one: $f(b)-f(a) = f'(c) (b-a)$, where $c \in (a,b)$. How did you get your statement, about existance of $t_n >n$? $\endgroup$ – Shot May 6 '17 at 8:56
  • $\begingroup$ +Still not clear what to do in the case of unbounded f(x) if the $lim f(x)$ doesn't exist? Is it still allowed to use l'Hopital rule? I didn't read about that, I thought limits in numerator and denominator should strictly exist?? $\endgroup$ – Shot May 6 '17 at 8:59

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