Solution verification of: $f$ periodic and $\lim_{x \to \infty} f(x)$ exists implies $f$ is constant 
Let $f$ be a periodic function. Show that if $\lim_{x \to \infty} f(x)$ exists in $\mathbb{R}$, then $f$ is a constant function. Use this result to prove that $\lim_{x \to \infty} \sin x$ doesn't exist.

My proof: let $\epsilon>0$. By hypothesis $\lim_{x \to \infty} f(x)$ exists, hence there exists $M_{\epsilon/2}>0$ such that $x \ge M_{\epsilon/2} \implies |f(x)-l|<\frac{1}{2}\epsilon$.
Since this holds for any $x \ge M_{\epsilon/2}$, given two arbitrary $x_1,x_2 \ge M_{\epsilon/2}$ it is $|f(x_1)-f(x_2)|=|f(x_1)-l+l-f(x_2)|\le|f(x_1)-l|+|f(x_2)-l|<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon=\epsilon$.
Since $\epsilon>0$ is arbitrary and $x_1,x_2 \ge M_{\epsilon/2}$ are arbitrary, this holds for any $\epsilon>0$ and for any $x_1,x_2 \ge M_{\epsilon/2}$; so $|f(x_1)-f(x_2)|=0$ for any $x_1,x_2 \ge M_{\epsilon/2}$, that is $f(x_1)=f(x_2)$ for any $x_1,x_2 \ge M_{\epsilon/2}$, thus $f$ is constant in $[M_{\epsilon/2},\infty)$.
By hypothesis $f$ is periodic with period $T>0$, being $[M_{\epsilon/2},M_{\epsilon/2}+T] \subset [M_{\epsilon/2},\infty)$ it follows that, in particular, $f$ is constant in $[M_{\epsilon/2},M_{\epsilon/2}+T]$. So, by periodicity, $f$ is constant in $\mathbb{R}$.
To prove that $\lim_{x \to \infty} \sin x$ doesn't exist I used the fact that $p \implies q$ is equivalent to $\neg q \implies \neg p$, so if $f$ is periodic and not constant, then $\lim_{x \to \infty} f(x)$ doesn't exist. Since $\sin x$ is periodic and not constant, $\lim_{x \to \infty} \sin x$ doesn't exist.
Is this all correct? I am not quite sure about the proof of the fact that $f$ is constant.
 A: The following paragraph is not correct:

"Since $\epsilon>0$ is arbitrary and $x_1,x_2 \ge M_{\epsilon/2}$ are arbitrary, this holds for any $\epsilon>0$ and for any $x_1,x_2 \ge M_{\epsilon/2}$, so $|f(x_1)-f(x_2)|=0$ for any $x_1,x_2 \ge M_{\epsilon/2}$, that is $f(x_1)=f(x_2)$ for any $x_1,x_2 \ge M_{\epsilon/2}$, thus $f$ is constant in $[M_{\epsilon/2},\infty)$."

Indeed, your $x_1, x_2$ depend on $\varepsilon$ such that you do not have, for $x_1, x_2$ fixed, $|f(x_1) - f(x_2)| < \varepsilon$ for all $\varepsilon$. To prove that $f$ is constant you must use the periodicity of the function $f$. Let us do it by contradiction.
Assume that $f$ is not constant, i.e. there is $x, y$ such that $|f(x) - f(y)| = \delta > 0$. If $T$ is the period of $f$, we may assume that $x, y \in [0,T]$ without lost of generality. Let $\varepsilon = \delta /2$. As you said, there is a $M_\varepsilon$ and $m \in \mathbb N$ such that $x + mT, y + mT \ge M_\varepsilon$ and
$$|f(x + m T) - f(y + mT)| = |f(x ) - f(y)| < \varepsilon.$$
This is a contradiction by definition of $x, y$.
A: If $f$ is not a constant, there are points $x_1,x_2\in[0,T)$ such that $f(x_1)\neq f(x_2)$. In particular $\lim_{n\rightarrow\infty}f(x_j+nT)=f(x_j)$ for $j=1,2$. Thus $\lim_{x\rightarrow\infty}f(x)$ does not exists.
