I was trying to prove this and I did a very similar argument as the one in this answer: What does $\lim\limits_{x \to \infty} f(x) = 1$ say about $\lim\limits_{x \to \infty} f'(x)$?

Basically, $\lim\limits_{x \to \infty} f'(x) \neq 0$ would imply that $f$ is not bounded, so its limit is not finite.

However, the theorem I was trying to prove makes the hypothesis that $f''(x)$ is bounded. Is this hypothesis necessary? It's sort of strange, because using just the argument I used would mean that (considering $f$ in n times differentiable) has all its derivatives equal to $0$.


marked as duplicate by Paramanand Singh, zarathustra, Claude Leibovici, Willie Wong, Newb Mar 19 '15 at 8:44

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Basically, $\lim_{x\to\infty}f'(x)\not=0$ would imply that f is not bounded, so its limit is not finite.


$$\lim_{x\to\infty}\frac{\sin(x^2)}{x} = 0$$ $$\lim_{x\to\infty}\frac{2x^2\cos(x^2) - 2x\sin(x^2)}{x^2} \text{ DNE}$$

Note that your original question did not presume that $\lim_{x\to\infty}f'(x)$ exist.

  • $\begingroup$ @robjohn We are considering $f(x)$ such that $\lim_{x\to\infty} f(x)$ is finite. $f(x)=\sin(x)$ fails to satisfy this. $\endgroup$ – StevenClontz Mar 19 '15 at 3:06
  • $\begingroup$ I was finding a contradiction to the original question, by showing an example where $\lim_{x\to\infty} f(x)$ is a finite real number, but $\lim_{x\to\infty} f'(x)\not=0$. $\endgroup$ – StevenClontz Mar 19 '15 at 3:30
  • $\begingroup$ Sorry, I misread the question. $\endgroup$ – robjohn Mar 19 '15 at 4:26

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