Q:

formal proof from calulus

Given $f:R \to R$, $f$ is differentiable on $R$ and $\lim_{x \to \infty}(f(x)-f(-x))=0$.
I need to show that there is $x_0 \in R$ such that $f'(x_0)=0$
I am trying to prove it by contradiction .... so i assume there is no $x_0 \in R$ such that  $f'(x_0)=0$ this means this function is strictly monotonic because the derivative "respects" the mean value theorem so it cannot be negative and positive without passing a zero value .... 
(from what i understand the derivative function of a continuous/differentiable functions isn't necessarily continuous but it still respect the mean value theorem i think it's called darboux's theorem )
now from $\lim_{x \to \infty}(f(x)-f(-x))=0$ ,we get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$
now the last step is the contradiction its very intuitive and even maybe obvious that we can't get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$ for strictly monotonic function but i cant think of a formal way to prove it . 

A:

Hint: Consider the function $g(x)=f(x)-f(-x)$. Since $g(0)=0$ and $\lim\limits_{x\to\infty}g(x)=0$, $g$ is either identically $0$ or it has a local maximum or minimum at $x_0\in(0,\infty)$. In the same way it was shown for Rolle's Theorem, we have that $g'(x_0)=0$. Thus,
$$
f'(x_0)+f'(-x_0)=0
$$
If $f'$ is not identically $0$ and $x_0\ne0$, apply the Intermediate Value Theorem.

