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This question already has an answer here:

Now I'm interesting in the question as follow,

Let $f\in C^1(\mathbb{R})$ and $f$ is bounded, $\lim_{x\rightarrow+\infty}f'(x)=0$, then $\lim_{x\rightarrow+\infty}f(x)$ exists.

Is this statement true? If not, please give a counterexample!

My try: if we remove the boundness condition of the function $f$, I can give a counterexample on the half line, namely, $f(x):=\ln x$.

Any answer will be appreciated!

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marked as duplicate by Siminore, Najib Idrissi, Clement C., Yiorgos S. Smyrlis, M Turgeon Mar 11 '14 at 12:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Your counterexample is not valid, as $\lim_{x\rightarrow+\infty}\log x=+\infty$. $\endgroup$ – alex Mar 11 '14 at 12:39
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My recipe for creating counterexamples of this sort is to design a function that oscillates between $-1$ and $1$.

In this case, you want the derivative to converge to $0$. This is easy to arrange piecewise: start at $0$ and have the function increase to $1$, then decrease to $-1$ more gradually, then increase to $1$ even more gradually, then decrease to $-1$ even more gradually....

Then you need to insert smooth arcs to replace the sharp corners, and voila, we have a bounded divergent function whose derivative converges to zero.

Usually once I go through this procedure, I can think up with a variation on the $\sin$ function that shares the behavior; pick any function $f(x)$ that converges to $+\infty$, but $f'(x)$ is strictly decreasing to zero. Then,

$$ \sin f(x) $$

will be a counterexample.

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Counterexample: $f(x) = \sin(\ln(x))$

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No. Take, for example the function

$$f(x)=\cos(\log x )$$

for $x>0$. This function is crearly $C^\infty$ and bounded in its domain. We have:

$$f'(x)=-\frac1{x}\sin\left(\log x \right)$$

which vanishes at infinity. However, since $\log x\rightarrow\infty$ when $x\rightarrow\infty$, $f(x)$ has no limit at infinity.

Just in case os you need a proof of that, take the sequence $\lbrace f(e^{\pi n}) \rbrace_{n=0}^\infty$, which is in fact $(1,-1,1,-1,\ldots)$.

I apologize if formulas are not correctly displayed. MathJax seems not to work well in this site for my computer.

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  • $\begingroup$ Sign error in $f'$. $\cos' = -\sin$... $\endgroup$ – AlexR Mar 11 '14 at 12:28

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