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.
Best Answer
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.