Prove that a function is not - one to one. Let $f : \mathbb{R}  \to \mathbb{R}$  be differentiable all over   $\; \mathbb{R}$ .
Also, $\;\;\lim_\limits{x \to \infty}f(x) = \lim_\limits{x \to -\infty}f(x) = \infty$
prove that there exists $c\in\mathbb{R}$ such that $f'(c) = 0$.
First of all, I sincerely apologize for my poor formatting and usage of Latex.
Now to the point. 
I want to prove that statement by showing that $f$ is not one-to-one and therefore show that there are $a,b \in\mathbb{R}$ such that $a\neq b$ and $f(a) = f(b)$ and afterwards use Rolle's theorem and finish.
How could I show that?
 A: Since $f(x)$ is differentiable over $R$, it follows that $f(x)$ is continuous over $R$; and so, because of this and the two limits given, we can find two real numbers $a, b$ such that $a \neq b$ and $f(a) = f(b)$ (and so, f is not 1-1 as you wanted to show). Now, finish off the problem by applying the Mean Value Theorem.  
A: Let $M>0$. By the two limits $\exists c_1,c_2>0:f(x)>M, x\in (-\infty, -c_2]\cup [c_1,\infty)$. Suppose $f(c_1)\neq f(-c_2)$ (otherwise the problem has finished) and furthermore $f(c_1)\gt f(-c_2)$ without loss of generality. By applying again the definition of the second limit, for the positive $f(c_1)$ exists a $c_2'>0$, such that $f(x)\gt f(c_1), x\leq -c_2'$. Provided immediately that $f(-c_2')>f(c_1)$, the Intermediate Value Theorem gives us for $f(c_1)\in [f(-c_2),f(-c_2')]$ a $ξ\in [-c_2',-c_2]$, such that $f(ξ)=f(c_1)$, thus the Rolle' s Theorem in $[ξ,c_1]$ completes the proof.
It is absolutely obvious that we found $ξ \neq c_1$ with $f(ξ)=f(c_1)$, so $f$ is not one to one.
