Extending Rolle's Theorem to Infinity Let $f : \mathbb{R} \to \mathbb{R}$ be a function which is differentiable everywhere on $\mathbb{R}$. If we know that 
$$\lim_{x\to-\infty}f(x) = L = \lim_{x\to\infty}f(x)$$
How would one go about showing that there must exist $c \in \mathbb{R}$, such that $f'(c) = 0$ for all cases at once?
 A: Consider
$$
g(x)=f(\tan x)
$$
defined over $(-\pi/2,\pi/2)$. Then we can extend it to $[-\pi/2,\pi/2]$ by setting
$$
g(-\pi/2)=g(\pi/2)=L
$$
The derivative of $g$ exists over $(-\pi/2,\pi/2)$ and is equal to
$$
g'(x)=\frac{f'(\tan x)}{\cos^2x}
$$
By Rolle's theorem, there exists $c$ such that $g'(c)=0$, so that $f'(\tan c)=0$.
You can do similarly for a function $f$ defined over $[a,\infty)$, differentiable over $(a,\infty)$ and such that
$$
f(a)=\lim_{x\to\infty}f(x)
$$
Define $g(x)=f(\tan x)$ over $[\arctan a,\pi/2)$ and the proof will be the same as before.
A: Derivatives satisfy the intermediate value property.
If $f'(c) \not= 0$ for all $c$, then either $f'(c) > 0$ for all $c$ or $f'(c) < 0$ for all $c$.
In either case, $f$ is strictly monotone ruling out $\displaystyle \lim_{x \to -\infty} f(x) = \lim_{x \to \infty} f(x)$.
A: If $f=L$ everywhere, the conclusion is clear. So let's assume $f(x_0)\ne L$ for some $x_0.$ WLOG, $f(x_0)>L.$ Because of the given limit conditions, there exists $a>|x_0|$ such that both $f(-a),f(a) < f(x_0).$ Then the maximum value of $f$ on $[-a,a]$ can't occur at either of the end points. It thus occurs at an interior point $c,$ and at that $c$ we have $ f'(c)=0.$
