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Let $f$ be a twice differentiable function such that $\lim_{x \rightarrow \infty} f(x)$ is finite and $\lim_{x \rightarrow \infty} f'(x) = 0$. Is it possible that $\lim_{x \rightarrow \infty} f''(x) \neq 0$?

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By $\lim_{x \rightarrow \infty} f''(x) \neq 0$, do you mean that the limit exists and is not $0$, or is the limit allowed to not exist? –  Chris Eagle Jul 1 '13 at 9:53
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I understood the second option, otherwise it is impossible: if $f''$ has a limit which is not $0$, then $f'$ cannot have a limit. –  Denis Jul 1 '13 at 9:56

1 Answer 1

It seems to work to take $f'(x)=\sin(x^2)/x$. Then it has limit $0$, and $f''(x)=\frac{2x^2\cos(x^2)-\sin(x^2)}{x^2}$, which has no limit in infinity.

As for $f(x)$, according to Wolfram Alpha it is $Si(x^2)/2$ where $Si$ is the sinus integral, so it has a finite limit which is $\pi/4$.

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