# If $f$ is differentiable on $[a,\infty)$ and $\lim\limits_{x\rightarrow\infty}f(x)=f(a)$, then $f'$ vanishes at some point.

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 $$x>x_0$$, $$0 so that $$\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 $$f(x) provided that $$a\leq x\leq p$$. Since $$f(x)\rightarrow f(a)$$ as $$x\rightarrow\infty$$, there exists $$q>\eta$$ such that $$f(x) 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 $$f'(\xi)=0$$.

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

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.