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
 A: Let $g(x)=f\bigl(\frac 1x\bigr)$ for $x\in (0,1]$ and $g(0)=f(1)$. Then $g$ is continuous in $[0,1]$ and derivable in $(0,1)$. By Rolle's Theorem there exists $c\in (0,1)$ such that $g'(c)=0$, hence
$$0=g'(c)=-f'\Bigl(\frac 1c\Bigr)\frac 1{c^2}$$
thus if $b=\frac 1c$, then $f'(b)=0$ and $b\in (1,+\infty)$.
A: Outline:
You may, and should, assume that $f$ is non-constant.
Pick $a$ so that $f(a)\ne f(1)$ (if no such $a$ exists $f$ is a constant function). 
Choose a value $c$ that is in between $f(1)$ and $f(a)$.  Use the Intermediate Value Theorem to show there is a $x_1\in(1,a)$ with $f(x_1)=c$. 
Use the Intermediate Value Theorem and your limit condition to show there is  an $x_2\in(a,\infty)$ with $ f(x_2)=c$. 
Now you're set to apply Rolle's Theorem.
A: 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 $x = \tan(t) \in (1, \infty)$, $$f'(x) = 0$$
A: If $f$ is constant, the statement is trivial. If it is not, then we have some $x_0$ such that $f(x_0)\neq f(1)$.


*

*If $f(x_0) > f(1)$: Define the midpoint between $f(1)$ and $f(x_0)$ as $y=\frac{f(x_0)+f(1)}{2}$. Because the function has values $f(1)$ and $f(x_0)$ on the interval $[1,x_0]$, there exists such $a$ that $f(a) = y$. Because the limit of $f(x)$ as $x\rightarrow\infty$ is $f(1)$, there exists such an $M$ that $f(x) < y$ for all $x\geq M$. Because the function $f$ has values $f(x_0)$ and $f(M)<y$ on the interval $[x_0, M]$, there exists such a point $b$ that $f(b)=y$. Now use Rolle's theorem on $[a,b]$.

*If $f(x_0) < f(1)$: Define the function $g(x) = -f(x)$ and use the previous argument on $g$.
A: Think about where $f(1)$ is. Pick a spot, any spot. Then think about all the points afterward. It's continuous and differentiable literally everywhere but possibly $f(1)$, and so we need to think about how it could not be flat somewhere.
First, we know it would be flat everywhere if $f(x)=f(1)$ a constant.
Next, we draw a line between $x = 1$ and $\infty$. At infinity, we have $f(\infty)=f(1)=K$ some finite constant. Then we think about how we draw a line from 1 to $\infty$ . If it's flat, ie $f(x)=f(1)=K$, then it is obviously true that $f'(x)=0$ everywhere between $1$ and $\infty$.
Now consider the case where it is non-constant.
Then we know there is a max (or min, by symmetry) $m$ such that $|f(m)|=|M|>k = f(1)$ on the number line. Now since $f$ is differentiable at $m \in (1,\infty)$, and $M$ is a max/min, it must be smooth and both sides smoothly approach it's flatness. This $m$ is your $b$ you are seeking to prove exists.
