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.
A: Although I feel that robjohn's answer is much simpler and better, I am providing an alternative proof.
Suppose that $f'(x) \neq 0$ for all $x \in \mathbb{R}$. Then since derivatives possess intermediate value property, $f'(x)$ is of constant sign. Let's take the case when $f'(x) > 0$ for all $x$ (the case $f'(x) < 0$ can be handled similarly). Clearly $f(x)$ is increasing everywhere and hence either $\lim\limits_{x \to \infty}f(x) = L$ or $\lim\limits_{x \to \infty}f(x) = \infty$. Also now $f(-x)$ is decreasing and hence either $\lim\limits_{x \to \infty}f(-x) = M$ or $\lim\limits_{x \to \infty}f(-x) = -\infty$. Considering each alternative we can see that we have the following two possibilities:


*

*$\lim\limits_{x \to \infty}\{f(x) - f(-x)\} = L - M > 0$

*$\lim\limits_{x \to \infty}\{f(x) - f(-x)\} = \infty$


But this contradicts the fact $\lim_{x \to \infty}\{f(x) - f(-x)\} = 0$. Note that in the above we must have $L > M$ because $f(x)$ is strictly increasing ($\lim\limits_{x \to \infty}f(-x) = M \leq f(0) < f(1) \leq L = \lim_{x \to \infty}f(x)$).
Update: From OP's comment below I guess further explanation is needed which I provide below. First I would advise OP to convince himself of the following result:

If $f(x)$ is increasing on interval $[a, \infty)$ then either $\lim\limits_{x \to \infty}f(x)$ exists or $f(x) \to \infty$ as $x \to \infty$.

There is a counterpart for decreasing functions:

If $f(x)$ is decreasing on interval $[a, \infty)$ then either $\lim\limits_{x \to \infty}f(x)$ exists or $f(x) \to -\infty$ as $x \to \infty$.

You can see the following examples $f(x) = -1/x, \arctan(x)$ for the increasing case with existing limits and $f(x) = \log x, e^{x}$ for diverging to $\infty$. Their negatives would serve as examples for the decreasing case. The function $f(x) = \cos x$ mentioned in comments by Hurkyl is not increasing and hence the above result does not apply to it. Once you are convinced to some extent by the above examples it is time to prove the result. I will establish the case for increasing function. So here goes the proof.
Let $f(x)$ be increasing on $[a, \infty)$. Clearly we can have only two possibilities: either $f$ is bounded on $[a, \infty)$ in which case $\sup_{x \in [a, \infty)} f(x) = A$ exists and is finite or $f$ is unbounded on $[a, \infty)$.
If $f$ is bounded I will show that $\lim_{x \to \infty}f(x) = A$. Clearly since $A = \sup f(x)$ for any given $\epsilon > 0$ there is a number $x_{0} \in [a, \infty)$ such that $A - \epsilon < f(x_{0}) \leq A$. Since $f$ is increasing therefore if $x > x_{0}$ we get $A - \epsilon < f(x_{0}) \leq f(x) \leq A$. It follows that we have $$A - \epsilon < f(x) < A + \epsilon$$ for all $x > x_{0}$ and hence $\lim_{x \to \infty}f(x) = A$.
If $f$ is unbounded then for any given positive number $N$ we have a value $x_{0} \in [a, \infty)$ such that $f(x_{0}) > N$. Again if $x > x_{0}$ we get $f(x) \geq f(x_{0}) > N$ and thus we have $f(x) > N$ for all $x > x_{0}$. It follows that $f(x) \to \infty$ as $x \to \infty$.
There is another issue which OP is facing and that is regarding split of limits in case of expressions like $f(x) \pm g(x)$ when $x \to \infty$. Clearly the rule $$\lim_{x \to \infty}\{f(x) \pm g(x)\} = \lim_{x \to \infty}f(x) \pm \lim_{x \to \infty}g(x)$$ holds when both the limits $\lim_{x \to \infty}f(x)$ and $\lim_{x \to \infty}g(x)$ exist. What happens when either one or both of them don't exist? In case one of them exists (say $\lim_{x \to \infty}g(x)$) then the behavior of $\{f(x) \pm g(x)\}$ as $x \to \infty$ is same as that of $f(x)$ as $x \to \infty$. When both the limits don't exist then we have only one case when we can say with guarantee and that is:
If $f(x) \to \infty$ and $g(x) \to \infty$ then $f(x) + g(x) \to \infty$ as $x \to \infty$.
Similar remarks can be made when the limits are $-\infty$.
