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Suppose $f$ is a real differentiable function on $[a,\infty)$. Prove that if $\lim\limits_{x\rightarrow\infty}f(x)=f(a)$, then $f'(\xi)=0$ for some $\xi>a$.

Proof   Suppose $f'$ vanishes at no point of $[a,\infty)$. Then, by Darboux's theorem, $f'$ does not change sign on $[a,\infty)$. Suppose, without loss of generality, $f'>0$. Then $f$ is strictly increasing on $[a,\infty)$. It follows that for any $x_0>a$ and $x>x_0$, $$0<f(x_0)-f(a)<f(x)-f(a)\mbox{,}$$ so that $\lim\limits_{x\rightarrow\infty}f(x)\not=f(a)$. Hence $f'(\xi)=0$ for some $\xi>a$.

Is this correct? Can someone provide a proof without the use of Darboux's theorem?


Thanks to copper.hat's comment.

If $f$ is constant then there is nothing to prove.
Suppose WLOG that $f(\eta)>f(a)$. By continuity of $f$, there exists $\eta>p>a$ such that $$f(x)<f(\eta)$$ provided that $a\leq x\leq p$. Since $f(x)\rightarrow f(a)$ as $x\rightarrow\infty$, there exists $q>\eta$ such that $$f(x)<f(\eta)$$ provied that $x\geq q$.
Since $[p,q]$ is compact, $f$ attains its maximum on $[p,q]$ at some point $\xi\in[p,q]$. Evidently, $f(\xi)=\sup f$, so that $f'(\xi)=0$.

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  • $\begingroup$ It must have a $\max $ or $\min$ somewhere. $\endgroup$
    – copper.hat
    Dec 14, 2021 at 6:22

1 Answer 1

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Your proof is fine. Here is one which is not using the Darboux theorem.

Note that $f$ is continuous.

If $f$ is constant we are done. Otherwise there is $b>a$ such that $f(b) \neq f(a)$, wlog $f(b) > f(a)$. Since $f$ tends to $ f(a)$ there is some $c>b$ such that $f(a) < f(c) < f(b)$ By the mean value theorem for continuous functions there is $d \in (a,b) $ such that $f(d) = f(c)$. Now the claim follows from Rolles Theorem.

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  • $\begingroup$ Nice approach. Thank you Sir. $\endgroup$ Dec 14, 2021 at 6:59

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