# $\lim f'(x) = l$

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 ?

• 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 $l=0$. Jan 8, 2021 at 2:58

## 1 Answer

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 $$l=0$$.

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