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Suppose $f(x)$ is continuous on $[1,+\infty)$, differentiable on $(1,+\infty)$. If $f(x)$ is bounded on $[1,+\infty)$ and has finite $\lim_{x\rightarrow \infty} f'(x)$, then it has finite $\lim_{x\rightarrow \infty} f(x)$

I know it's false, but don't see why: if the function is bounded, then it either has finite limit or doesn't have limit at all, i.e. oscillating. Having derivative, which limit is finite, means that as we go to infinity our function turns into monotonic one. Bounded monotone function should have a finite limit.

Correct?

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  • $\begingroup$ How do you know this : >that as we go to infinity our function turns into monotonic one. $\endgroup$
    – dajoker
    Jun 29, 2013 at 10:43
  • $\begingroup$ What about $f(x)=\sin(\frac{1}{n}x)$? $\endgroup$
    – Hui Yu
    Jun 29, 2013 at 10:48
  • $\begingroup$ @HuiYu It's unclear what the $1/n$ does to make that different from just $\sin(x).$ $\endgroup$
    – coffeemath
    Jun 29, 2013 at 10:54
  • $\begingroup$ Note (I think) that bounded $f$ would imply that if $f'$ has a limit, that limit must be $0$. $\endgroup$
    – coffeemath
    Jun 29, 2013 at 11:15

1 Answer 1

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$f(x)=\sin(\sqrt{x})$ has $f'(x)=\cos(\sqrt{x})/(2\sqrt{x})\to 0$.

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  • $\begingroup$ If this is graphed, one sees it gets flatter and flatter as $x$ increases, the sine arcs getting widened out. But even a small derivative can eventually let one reach some fixed height like 1. $\endgroup$
    – coffeemath
    Jun 29, 2013 at 11:28

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