Rolle theorem on infinite interval We have :
$f(x)$ is continuous  on $[1,\infty]$ and differentiable on $(1,\infty)$
$\lim\limits_{x \to \infty}f(x) = f(1)$
we have to prove that : there is $b\in(1,\infty)$ such that $f'(b) = 0$
I'm sure we have to use Rolle's theorem so, I tried using Mean Value theorem and using the limit definition  at $\infty$
Any ideas how can I use them ?
Update :
after seeking the answers that I've got :
I'm having trouble finding $\boldsymbol x_{\boldsymbol 1}\neq\boldsymbol x_{\boldsymbol 2}$ such that $\boldsymbol{f(x}_{\boldsymbol 1}\boldsymbol {)=f(x}_{\boldsymbol 2}\boldsymbol )$
*I need a formal solution

An alternative to the existing answers:
Because of your hypotheses, the function defined by: $$g(t) = \begin{cases} f(\tan(t)) & t \in [\pi/4, \pi/2) \\ f(1) = \lim_{x \to \infty} f(x) & t = \pi/2 \end{cases}$$ is continuous on $[\pi/4, \pi/2]$, differentiable on $(\pi/4, \pi/2)$, and $g(\pi/4) = f(1) = g(\pi/2)$. Apply Rolle's theorem to get $t \in [\pi/4, \pi/2]$ such that $g'(t) = 0$.
Compute $g'(t) = \tan'(t) f'(\tan(t)) = (1 + \tan(t)^2) f'(\tan(t)) = 0$. Therefore for $x = \tan(t) \in (1, \infty)$, $$f'(x) = 0$$
