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.
 A: Assuming that $f$ is non constant (otherwise the statement is obviously false), let $T$ be a period and suppose $a,b\in(0,T)$ such that $f(a)\ne f(b)$, which exist because $f$ is non constant.
Consider the sequences $a_n=a+nT$, $b_n=b+nT$. Compute
$$
\lim_{n\to\infty}f(a_n),
\qquad
\lim_{n\to\infty}f(b_n)
$$
and conclude.
A: Suppose $\lim_{x \to \infty} f(x) = L$. 
For any $x_0$ with $f(x_0) \neq L$, let $\epsilon = |f(x_0) - L|/2$. 
There exists $K$ such that if $x > K$ we have $|f(x) - L| < \epsilon$.  Choose $k$ such that $x_0 +kp > K.$
Then we arrive at a contradiction
$$|f(x_0) - L| = |f(x_0 + kp) - L| < \epsilon = |f(x_0) - L|/2$$
A: You are on the right track but You aren't using the idea that L is the limit or that $|f(x) - L| < \epsilon$.  Your triangle inequality is legit and true for any $L = f(x_0)$ whether a limit or not and isn't, therefore, in anyway a contradiction.
More to the point: if $f$ isn't constant then there are $x, y$ s.t $f(x) \ne f(y)$ and if $f$ has a period of $p$ then $f(x + np) \ne f(y+ np)$ for all $n \in \mathbb N$.  So for any $N \in \mathbb R$ there are $x+np, y + np > N$ such that $|f(x + np) - f(y + np)| = |f(x) - f(y)| = c_1 > 0$.
This is enough. 
For any $L$, $c_1 = |f(x+np) - f(y+np)| \le |f(x + np) - L| + |f(y + np) - L|$.  So either $|f(x + np) - L| \ge c_1/2$ or $|f(y+ np) - L| \ge c_1/2$. 
So if $\epsilon = c_1/2 > 0$ there is no $N$ or $L$ such that $w > N$ implies $|f(w) - L| < \epsilon$ (as $w$ can always equal either $x+np$ or $y+np$ for some $n$). 
So there is no limit.
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Oh, I really like RRL's answer!  Inspired by it, I'm modifying mine.  (Same idea but a direct proof.)
Let $L$ be any value.  As $f$ is not constant there is an $x$ such that $f(x) \ne L$.  Let $c = |f(x) - L| > 0$.  As $f$ is periodic, let's say with a period $p$, $f(x + np) = f(x)$.  For any $N$ there is an $x + np > N$ where $|f(x+np) - L|= c$.  So there is no $N$ where $w > N$ implies $|f(w) - L| < c$.
So no value $L$ can be a limit of $f$.
A: Your argument doesn't work because your computation only gives you $|f(x)-L|\leq |f(x)-L|$, not $|f(x)-L|<|f(x)-L|$.  What you want to do is use the fact that $f(x)=f(x+np)$ for all $n\in\mathbb{Z}$, and for any $x$, you can choose $n$ large enough so that $x+np>\delta$.  You should then be able to conclude from this that actually $f(x)=L$ for all $x$, so $f$ is constant.
A: 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.
