# $\lim_{x\to \infty}f(x)$ doesn't exist. Show that there is $x_0\in\mathbb{R}$ such that $f'(x_0)=0$.

Let $f: \mathbb{R} \to \mathbb{R}$ such that:

(i) $f$ is differentiable in $\mathbb{R}$

(ii) The limit $\lim_{x\to \infty}f(x)$ doesn't exist (EDIT : neither finite nor infinite).

Show that there is $x_0\in\mathbb{R}$ such that $f'(x_0)=0$.

I understand intuitively that these terms satisfy that the function is not injective and thus I can use lagrange, but I'm wondering how to show it formally.

Any help would be appreciated

If $f'$ has no zeros, it follows from Darboux's theorem that either $f'(x)$ is always greater than $0$ or always smaller. Therefore, $f$ is either increasing or decreasing. But then $\lim_{x\to\infty}f(x)$ must exist (in $\mathbb{R}\cup\{+\infty,-\infty\}$).

• Exactly what I was looking for. Thank you! – Moshe King Jan 25 '18 at 17:12
• Just a small question. If the function is increasing, why does that imply there's a limit in infinity? @José Carlos Santos – Moshe King Jan 25 '18 at 17:15
• @GADI Because either the function is unbounded above, in which case $\lim_{x\to+\infty}f(x)=+\infty$, or it is, in which case $\lim_{x\to+\infty}f(x)=\sup\{f(x)\,|\,x\in\mathbb{R}\}$. – José Carlos Santos Jan 25 '18 at 17:17

$$f(x) = x$$

• Differentiable function

• the limit $\lim_{x \to \infty} f(x)$ doesn't exist.

There is no such $x_ 0 ∈ \Bbb R$ such that $f′(x_0)=0$.

• Based on the context, I'm guessing (almost sure...) that $\lim_{x \to \infty} f(x) = \infty$ counts as the limit existing for the asker. – user296602 Jan 25 '18 at 17:07