Prove that there exists $c \in \mathbb{R}$ such that $f'(c)=0$ if $ \lim_{x\to-\infty}f(x) =\lim_{x\to\infty}f(x) = 0 $.? Suppose that $f$  is differentiable on $ \mathbb{R} $ and $ \lim_{x\to-\infty}f(x) =\lim_{x\to\infty}f(x) = 0 $.
Prove that there exists $c \in \mathbb{R}$ such that $f'(c)=0$ .
and can I extending this claim  to $\mathbb{C}$ ?
 A: If $f$ is zero everywhere, then it's trivially true. Else WLOG $f(x)>0$ for some $x\in\mathbb R$. The limit conditions imply $f$ has a global maximum by Weierstrass' theorem on sufficiently large domain. The Fermat's theorem then implies the derivative is $0$ there.
The complex case is false as is witnessed by $\dfrac{e^{it}}{t^2+1}$. (I'm assuming $f:\mathbb R\to\mathbb C$ is the generalization)
A: Define
$$
g(x)=\begin{cases}
0 & \text{if $x=-\pi/2$},\\[2ex]
f(\tan x) & \text{if $-\pi/2<x<\pi/2$},\\[2ex]
0 & \text{if $x=-\pi/2$}.
\end{cases}
$$
Then $g$ is continuous on $[-\pi/2,\pi/2]$ and differentiable on $(-\pi/2,\pi/2)$. Since $g(-\pi/2)=g(\pi/2)$, Rolle's theorem applies and there exists $k\in(-\pi/2,\pi/2)$ such that
$$
g'(k)=0.
$$
But $g'(x)=f'(\tan x)\cdot(1+\tan^2x)$ on $(-\pi/2,\pi/2)$, so
$$
f'(\tan k)=0
$$
and $c=\tan k$ is the number you're looking for.
In other words, you can apply Rolle's theorem also on $[-\infty,\infty]$.
It's not clear what generalization to $\mathbb{C}$ you have in mind.
