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Let $f: \Bbb R \rightarrow \Bbb R$ be differentiable with $f'$ uniformly continuous. Suppose $\lim\limits_{x\rightarrow \infty}f(x)=L$ for some $L$. Does $\lim\limits_{x\rightarrow \infty}f'(x)$ exist?

I have no idea about this problem. Could you please give a hint? Thank you very much for your help.

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Outline:

Without loss of generality, $L=0$ (if not, consider $g=f-L$).

Suppose by contradiction that $f'(x)\not\xrightarrow[n\to\infty]{}0$. Negating the definition, it means there exists $\varepsilon_0>0$ and a sequence $(x_n)_{n\geq 0}$ with $x_n\xrightarrow[n\to\infty]{} \infty$ such that $\lvert f'(x_n) \rvert > \varepsilon_0$ for all $n\geq 0$. By a standard argument (pick, out of the two, the subsequence with the most points; then wlog you can assume that the derivative is positive, otherwise consider $-f$ instead of $f$) we get a sequence $(y_n)_{n\geq 0}$ with $y_n\xrightarrow[n\to\infty]{} \infty$ such that $f'(y_n) > \varepsilon_0$

Let $\delta > 0$ be the uniform continuity modulus for $f'$ (which is assumed uniformly continuous) and $\varepsilon\stackrel{\rm def}{=}\frac{\varepsilon_0}{2}$. We get that for any $n\geq 0$, and any $x\in[y_n-\delta, y_n+\delta]$, we have $f'(x) \geq \varepsilon_0-\varepsilon = \frac{\varepsilon_0}{2}$. But then, you now something about $f$ on $[y_n-\delta, y_n+\delta]$: namely, that it "goes away" from $y_n$ at least at the rate of an affine function with slope $\frac{\varepsilon_0}{2}$.

Now use that to show that you must have infinitely many points $z_n$ (one around each $y_n$) such that $\lvert f(z_n) \rvert \geq \frac{\delta\varepsilon_0}{2}>0$. How does that contradict the assumption that $f(x)\xrightarrow[x\to\infty]{} 0$?

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  • $\begingroup$ Why does the limit of the derivative exist? I don't understand much your solution. Could you please explain more? $\endgroup$ – user52523 Nov 21 '16 at 13:21
  • $\begingroup$ Suppose it does not exist, or more precisely that $f'$ does not converge to zero. Then derive a contradiction (namely, that the limit of $f$ then cannot exist). This shows that we must have $\lim_\infty f' = 0$. $\endgroup$ – Clement C. Nov 21 '16 at 13:23
  • $\begingroup$ Could you explain more about "goes away" from ynyn at least at the rate of an affine function with slope $\epsilon_0/2$? I understand your idea, but this sentence is not clear to me. Thank you very much. $\endgroup$ – user52523 Nov 21 '16 at 13:31
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    $\begingroup$ Suppose (the negative case is similar) that $f'(x) > \gamma >0$ for all $x\in[a-\delta,a+\delta]$. Then, we get (e.g., by integrating) that $f(a+\delta) \geq f(a)+ \delta \gamma$, and $f(a-\delta) \leq f(a) - \delta \gamma$. (Compared to the affine function $h(x)=f(a)+\gamma(x-a)$ which has rate/slope $\gamma$, the function $f$ grows faster). $\endgroup$ – Clement C. Nov 21 '16 at 13:34
  • $\begingroup$ C. Thank you very much. I got it. $\endgroup$ – user52523 Nov 21 '16 at 14:31
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Hint: If the limit on f exists, it means f becomes constant as x goes to infinity. What can you say about the derivative of a constant function?

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    $\begingroup$ ?! This is completely false -- in particular, it misses all about the uniform continuity, which has to play a role. $f$ does not have to become a constant function at all. $f(x) = \frac{1}{x}\sin(x^2)$? $\endgroup$ – Clement C. Nov 21 '16 at 3:46

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