calculus - Prove there's $x_0$ such that $f'(x_0)=0$

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable at $\mathbb{R}$ and:

$$\lim_{x\rightarrow \infty}\left( f(x)-f(-x) \right) = 0$$

Show there's $x_0$ such that $f'(x_0) = 0$.

I tried to use the limit definition, but couldn't get much further.
I'll be glad for guidance.

Thanks.

Answer

If $f$ is a constant function, the result is easy.

Here is an outline of one way to prove such an $x_0$ exists if $f$ is not a constant function:

1. Choose $a, b$ with $f(a)< f(b)$ for some $a,b$.





2. Show that there is an $N>\max\{|a|,|b|\}$ so that either $\max\{f(N), f(-N)\}f(a)$(choose$N$so that$f(N)$is within$|f(b)-f(a)|/2$of$f(-N)$; drawing a picture will help here).




3. Use the result of 2. and the Intermediate Value Theorem to show $f(c)=f(d)$ for some $c\ne d$.




4. Wrap things up with Rolle's Theorem.

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Alternatively:

If $f'(x)$ is never $0$, then by Darboux's Theorem, $f'$ is either positive everywhere or negative everywhere). In this case $f$ is either strictly increasing or strictly decreasing. But then, for $x>1$, $|f(x)-f(-x)|> |f(1)-f(-1)|>0$; whence $\lim\limits_{x\rightarrow\infty} \bigl(f(x)-f(-x)\bigr)\ne0$.