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Prove that if $\lim\limits_{x\to\infty}f(x)$ and $\lim\limits_{x\to\infty}f''(x)$ exist, then $\lim\limits_{x\to\infty}f'(x)=0$.

I can prove that $\lim\limits_{x\to\infty}f''(x)=0$. Otherwise $f'(x)$ goes to infinity and $f(x)$ goes to infinity, contradicting the fact that $\lim\limits_{x\to\infty}f(x)$ exists. I can also prove that if $\lim\limits_{x\to\infty}f'(x)$ exists, it must be 0. So it remains to prove that $\lim\limits_{x\to\infty}f'(x)$ exists. I'm stuck at this point.

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This is similar to a recent Putnam problem, actually. By Taylor's theorem with error term, we know that for any $x$, $$ f(x+1) = f(x) + f'(x) + \tfrac12f''(t) $$ for some $x\le t\le x+1$. Solve for $f'(x)$ and take limits....

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  • $\begingroup$ Nice! But there should be a solution without Taylor's theorem, because it has not been introduced in this part of the book. $\endgroup$ – user19033 Nov 7 '11 at 6:41
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Hint $\ $ This follows easily from L'Hôpital's rule since

$$\rm \lim_{x\to\infty}\ f-f'\ =\ \lim_{x\to\infty}\frac{e^x\ (f-f')}{e^x}\ =\ \lim_{x\to\infty}\frac{e^x\ (\:f-f'+f'-f'')}{e^x}\ =\ \lim_{x\to\infty}\ f-f''\ exists$$

See also the similar classic Hardy old problem.

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  • $\begingroup$ Very clever one! $\endgroup$ – Pedro Tamaroff Apr 11 '12 at 19:03

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