Does $\lim\limits_{x\rightarrow \infty}f'(x)$ exist? 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.
 A: 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 $f(x)\xrightarrow[x\to\infty]{} 0$?
A: 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?
