If $f:\mathbb{R}\mapsto\mathbb{R}$ is a non-constant periodic function than $\lim_{x \to \infty} f(x)$ does not exist I am trying to prove if $f:\mathbb{R} \mapsto \mathbb{R}$ is any non-constant periodic function than $\lim_{x \to \infty} f(x)$ does not exist.  What I have so far is this:
Suppose $f$ is periodic with period $p$.  Suppose the limit exists and is equal to $L$.  Than for every $\epsilon > 0$ there exists $\delta > 0$ such that $x > \delta \implies |f(x) - L| < \epsilon.$  Since $f$ is periodic we have
$|f(x) - L| = |f(x) + f(x+p) - f(x+p) - L| = |(f(x) - f(x+p)) + (f(x+p) - L)| \leq |f(x) - f(x+p)| + |f(x+p) - L| = |f(x) - f(x)| + |f(x) - L| = |f(x) -L|$
Thus $|f(x) - L| < |f(x) - L|$ a contradiction.  Therefore $\lim_{x \to \infty} f(x)$ does not exist.

Your reasoning implies that $|f(x)-L|\le|f(x)-L|$, which it is not a contradiction. In fact, as comments suggest, the supossition that $f$ is periodic and has limit at infinity should imply that $f$ is constant.
So, let's suppose that $f$ is periodic and not constant. Then, there exist two real numbers $x,y$ such that $f(x)\neq f(y)$. Let $\epsilon=|f(x)-f(y)|/2$ and $p$ be the period. Suppose also that $L=\lim_{x\to\infty} f(x)$.
Now, for any $K\in\Bbb R$ we have some natural numbers $m,n$ such that $x+mp>K$ and $y+np>K$. You have to prove that $|f(x+mp)-L|<\epsilon$ and $|f(y+np)-L|<\epsilon$ together lead to a contradiction.
