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I know that if $f$ is continuously differentiable and $f'$ is uniformly continuous then we have $$ \exists\lim_{x\to\infty}f(x)\Rightarrow \lim_{x\to\infty} f'(x)=0. $$ But I was thinking of changing these hypotheses so that the conclusion is still true. I thought of the following: If $f:\mathbb R\to\mathbb R$ is differentiable, strictly decreasing and bounded below then $\displaystyle\lim_{x\to\infty} f'(x)=0.$

Geometrically it makes a lot of sense, since taking $X=\{f(n):n\in\mathbb N\}$, then exists $\inf(X)=c$. Therefore $$c=\lim_{n\in\mathbb N}f(n)=\lim_{x\to \infty}f(x).$$ Then there is a horizontal asymptote to the right of $f$, $y=c$, with slope $0$. So I can think of $\lim_{x\to \infty}f'(x)=0$.

But I feel that rigor is too soft. Can someone help me to give more mathematical rigor to this part?

Zaragosa
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    See https://math.stackexchange.com/a/788818/42969 for a counterexample, or this: https://math.stackexchange.com/a/863763/42969 – both found with a Google search for “decreasing bounded function derivative has limit zero” – Martin R Jan 28 '22 at 20:21
  • Thanks @MartinR , I hadn't thought of it that way. I managed to prove that if $f$ is bounded, of class $C^1$ and $\displaystyle\lim_{x\to \infty}f'(x)=M$, then $M=0$. It is not so strong because I already suppose the existence of the limit but at least it is something I think. – Zaragosa Jan 28 '22 at 20:33

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