Why do non-constant periodic functions have no limit at infinity? Why do periodic functions (like $\cos$ or $\sin$ or $\tan$) have no limit at infinity?  
I can guess that it is because their values don't converge but repeat over and over, but I would like to know what is the formal analytical (that is with $\epsilon$'s and $\delta$'s) proof of the statement.
 A: The reasoning is quite simple: take any periodic function $f$ such that $\forall x\ \ \  f(x+T)=f(x)$ and such that $f$ takes at least two different values (i.e. $f$ is not constant). Two two such values $a$ and $b$.
Now suppose that $f$ has a limit $l$ for $x\to\infty$. Take $\varepsilon = |a-b|/4$, then there exists $R$ such that $$\forall x>R \quad |f(x)-l|<\varepsilon=|a-b|/4.$$
But there exists $y_a$ such that $f(y_a)=a$ and $y_b$ such that  $f(y_b)=b$, together with condition $y_a>R$ and $y_b>R$ (that's because $f$ is periodic). Put these values into the formula above and obtain a contradiction.
Edit
Why such $y_a$ and $y_b$ exist? We chose $a$ and $b$ to be the values of $f$, hence there exist $x_a\in \Bbb R$ such that $f(x_a)=a$. Now by periodicity you obtain that $$\forall n\in \Bbb Z\quad f(x_a+nT) = f(x_a)=a,$$
therefore if we chose $\Bbb N\ni m>\frac{R-x_a}{T}$, we obtain that $y_a=x_a+mT>R$ and $f(y_a)=a$. Similarly we can show the existence of $y_b$. 
A: Any periodic function $f:\mathbb R\rightarrow \mathbb R$ that assumes more than one value can have no limit at $\pm \infty$.
Indeed, suppose the limit at $+\infty$ is the real number $L$ (the proof is similar for $-\infty$, so I omit it here). This means that we may make $|f(x)-L|$ as small as desired by taking $x$ sufficiently large.
Now this is already problematic. Since $f$ assumes two distinct values $a$ and $b$, and $f$ is periodic, there are arbitrarily large values of $x$ for which $f(x)=a$, and there are arbitrarily large values of $x$ for which $f(x)=b$, yet $f(x)$ must be bounded away from at least one of $a$ or $b$ (both, in fact, if $L$ is not one of them) for sufficiently large $x$.
The essential idea is that as $x$ increases to $+\infty$, $f(x)$ must visit $a$ and $b$ over and over, but the existence of the limit requires that $f(x)$ must eventually stay close to $L$. The only way all of this can happen is if $a=b=L$ and $f$ is constant.
A: You need do define convergence in a precise manner and then work with it.

Def. Let $f \colon ℝ → ℝ$ be a function. Then $f$ converges at $∞$ if there is a point $y ∈ ℝ$ such that for any $ε > 0$ there is an $N ∈ ℕ$ such that $f$ takes $(N..∞)$ to $B_ε(y)$, i.e. $f(N..∞) ⊂ B_ε(y)$.
We then call $y ∈ ℝ$ the limit of $f$ and write $f(x) \overset{x → ∞}{\longrightarrow} y$ or $y = \lim_{x → ∞} f(x)$.

So $\sin$ does not converge at $∞$:

Let $y ∈ ℝ$ be candidate for the limit of $\sin$. Then choose $ε = 1$. Now either $-1 \notin B_ε(y)$ or $1 \notin B_ε(y)$ since $d(-1,1) = |-1-1| = 2 = 2ε$ and any two points in $B_ε(y)$ have distance less than $ε + ε$ (by triangle inequality). Preserving generality, let $1 \notin B_ε(y)$.
But for any $N ∈ ℕ$, choose $x = 2πN + π/2$. Then $\sin x = \sin (2πN + π/2) = \sin π/2 = 1$, so $\sin x \notin B_ε(y)$. Therefore, $\sin$ does not converge to $y$ at $∞$. As $y$ was arbitrary, $\sin$ doesn’t converge at $∞$ at all.

As you can see, this argument works for all periodic functions that assume at least two different points. You can also generalize the definition for limits if you work with general neighbourhoods. Experiment.
