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Let $f : \mathbb{R} \to \mathbb{R}$ be differentiable function.

If $\lim_{x\to \infty}f(x) = 0$ , then $\lim_{x \to \infty} f^{'}(x) $ exists?

I think the answer is No.

Because I think there may exists function $f$ alternanting its sign as $x \to \infty$.

Could you give me a such function or Prove above statement?

Michael Hardy
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Seongqjini
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1 Answers1

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I think the typical example in this case is something like $f(x) = \frac{\sin(x^3)}{x}$. Then by the squeeze theorem we have $\lim_{x\to \infty} f(x) = 0$ but $$f'(x) = \frac{3x^2\cos(x^3)\cdot x - \sin(x^3)}{x^2} = 3x\cos(x^3) + o(1)\,\,\,\,\, \text{ as } x \to \infty$$ so $\lim_{x\to \infty} f'(x)$ doesn't exist.

User8128
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  • The notation $f \sim g$ is mostly used to mean that $\lim \frac{f(x)}{g(x)} = 1$, but if $f$ and $g$ do not have the same constant sign after some sufficiently large $x$, we can't conclude that $\lim f(x)$ exists iff $\lim g(x)$ exists. Maybe it would be better to use the notation $\approx$ or something like that –  Aug 09 '16 at 18:00
  • I use it to mean 'behaves like' in some vague sense. You're right, I could be more precise. I will edit. – User8128 Aug 09 '16 at 19:54