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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 $0$.

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El Cid
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  • See the following questions http://math.stackexchange.com/questions/396707/if-fx-to-0-as-x-to-infty-and-f-is-bounded-show-that-fx-to0-as-x and http://math.stackexchange.com/questions/730411/behavior-of-fx-as-x-to-infty-given-the-behavior-of-fx-and-fx – Paramanand Singh Mar 19 '15 at 06:11
  • especially see the proof by J E Littlewood http://math.stackexchange.com/a/391864/ – Paramanand Singh Mar 19 '15 at 06:30

1 Answers1

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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.

Counterexample:

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

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

Steven Clontz
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