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Let $f : \left]0, +\infty\right[ \to \mathbb R$ a differentiable and bounded function such that $$\lim_{x \to +\infty} f'(x) = l$$ Show that $l = 0$.

My attempt is the following : Suppose that $l>0$, then $\forall \epsilon > 0, \exists x_0 > 0, \text{ such that } \forall x>x_0, \lvert f'(x) - l \rvert \leq \epsilon $. Taking $\epsilon = \frac{l}{2}$, we have that $\frac{l}{2} \leq f'(x)$ $\forall x > x_0$.

Then by the mean value theorem, we can say that $\forall x > x_0, f(x) \geq \frac{l}{2}(x-x_0) + f(x_0)$.

Hence, $\lim_{x \to +\infty}f(x)= +\infty$. I guess we can do the same for $l < 0$ and deduce that $\lim_{x \to +\infty}f(x)= -\infty$. So if $f$ is bounded then $l=0$.

Is it correct ? Is there any direct proof ?

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  • $\begingroup$ Your argument is fine. Another way is L'Hospital's Rule which says that in this case $f(x) /x\to l$ and $f$ is bounded so $l=0$. $\endgroup$
    – Paramanand Singh
    Jan 8, 2021 at 2:58

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That is correct. If $\lim_{x \to +\infty} f'(x)$ exists and is not zero then $f$ diverges to $+\infty$ or $-\infty$ for $x \to \infty$ and is not bounded.

As an alternative (direct) proof you can apply the mean-value theorem to intervals $[n, 2n]$ with positive integers $n$: $$ \frac{f(2n)-f(n)}{n} = f'(c_n) $$ with some $c_n \in (n, 2n)$. Then note that for $n\to \infty$, the left-hand side converges to zero (because $f$ is bounded) and the right-hand side converges to $\lim_{x \to +\infty} f'(x) = l$. It follows that $l=0$.

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  • $\begingroup$ Yes, MVT is the way to do it. Trying to integrate $f'$ requires more justification. $\endgroup$
    – GEdgar
    Jan 7, 2021 at 22:06

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