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

Question

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

Answer 1

$$f(x) = x$$

• Differentiable function

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

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

Contradiction.

Answer 2

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 $\mathbb{R}\cup\{+\infty,-\infty\}$).