Limit of $f(x)/x$ given $f$ has periodic derivative Suppose $f: \mathbb{R} \to \mathbb{R}$ has periodic derivative: that is, $f'(x+c)=f'(x)$ for some $c \in \mathbb{R}^+$ and all $x \in \mathbb{R}$. Show that $\lim_{x \to \infty} \frac{f(x)}{x}$ exists. 
My attempt: The period does not matter. Showing it for one $c \in \mathbb{R}^+$ shows it for any such with the same proof so you might as well assume period $1$ or $2\pi$. Trying a few examples, if $f(x)=\sin x$ then the limit is 0. If $f(x)=ax+\sin x$ then the limit is $a$. So it seems as if the limit should be the limit of $f'(x)/x$ as $x \to \infty$. I assume period $1$ for convenience. 
We have $f(x)$ differentiable on $[1,x+1]$, so that by the Mean Value Theorem, $f(x+1)-f(1)=f'(c)x$ for some $c \in [1,x+1]$. Dividing by $x$ brings us close to what we want. But all the algebraic manipulations I've tried using the Mean Value Theorem on the intervals $[1,x], [1,x+1],[x,x+1]$ have come up terribly short. Are these the right intervals to try or am I off with what the limit should be?
 A: because $f(x) = C+\int_0^x f'(t) dt$ and $f'(t+a) = f'(t)$ :
$$\int_0^{n a} f'(t) dt = n \int_0^a f'(t) dt$$
so that 
$$f(x) = C+\lfloor x / a \rfloor \int_0^a f'(t) dt + \int_0^{x-a \lfloor x / a \rfloor} f'(t) dt = x \int_0^a \frac{f'(t)}{a} dt + \mathcal{O}(1)$$
A: The key is to realizing that the map
$$ x\mapsto f(x+c)-f(x) $$
is constant, since the map has derivative $f'(x+c)-f'(x) = 0$. Let $C$ satisfy $f(x+c)-f(x) = C$ for all $x\in\mathbb{R}$. I claim that $\lim\limits_{x\rightarrow\infty}{\frac{f(x)}{x}} = \frac{C}{c}$.
To see this, let $M = \max\limits_{x\in[0,c]}{|f(x)|}$ (note that $M$ exists because $f$ is continuous). For each $x\in\mathbb{R}$, let $\bar{x}$ be the greatest integer multiple of $c$ less than $x$. We then have
\begin{align*} f(x) &= (f(x)-f(x-c)) + (f(x-c)-f(x-2c)) + \dots + (f(x-\bar{x}+c)-f(x-\bar{x})) + f(x-\bar{x}) \\
&= \frac{\bar{x}}{c}C + f(x-\bar{x}) 
\end{align*}
and hence
$$ \lim\limits_{x\rightarrow\infty}{\frac{f(x)}{x}} = \lim\limits_{x\rightarrow\infty}{\frac{\frac{\bar{x}}{c}C + f(x-\bar{x})}{\bar{x}}\frac{\bar{x}}{x}} = \lim\limits_{x\rightarrow\infty}{\left(\frac{C}{c} + \frac{f(x-\bar{x})}{\bar{x}}\right)\frac{\bar{x}}{x}}$$
Since $x-\bar{x}\in[0,c]$ for all $x$, we have $|f(x-\bar{x})|\le M$ for all $x$, and hence $\lim\limits_{x\rightarrow\infty}{\frac{f(x-\bar{x})}{\bar{x}}} = 0$. Since $\lim\limits_{x\rightarrow\infty}{\frac{\bar{x}}{x}} = 1$, the conclusion follows.
