Do I need knowledge of continuous functions in order to rigorously determine that $\lim_{x \to \infty}\sin(x)$ does not exist? In problem 33 iv Chapter 5 of Spivak's Calculus, there is a comment in the solution manual that (in essence) remarks $\lim_{x \to \infty} \sin x$ does not exist.
I wanted to see if I could prove this statement by negating the formal definition of $\lim_{x \to \infty}f(x)=L \iff \exists L \in \mathbb R \Big [ \forall \epsilon \gt 0 \ \exists N \in \mathbb R \text { s.t. } \forall x \in \mathbb R \big ( x \gt N \rightarrow \lvert f(x) - L \rvert \lt \epsilon \big ) \Big ]$
i.e. I wanted to write a proof that:

$\forall L \in \mathbb R \Big [\exists \epsilon \gt 0 \text{ s.t. } \forall N \in \mathbb R \ \exists x \in \mathbb R \text{ s.t. }\big (x \gt N \land \lvert \sin(x) - L \rvert \geq \epsilon \big) \Big]$

I'll write what I have so far and highlight the missing piece that I am finding difficult to navigate around. As an "FYI" for those unfamiliar with Spivak's Calculus, the subject of Continuous Functions is not found until chapter 6, and I suspect that my issue arises from trig functions being more..."tackle-able"...once continuity is rigorously defined.

Relevant Lemma: $\lvert \sin(x) \rvert \leq 1 \quad$ (I do not believe this statement, in and of itself, requires knowledge of continuous functions)
Consider two cases: $\lvert L \rvert \gt 1$ or $\lvert L \rvert \leq 1$.

Case 1: Note that $\lvert L \rvert \gt 1 \implies \frac{\lvert L \rvert -1}{2}\gt 0$
Let $0 \lt\epsilon \lt \frac{\lvert L \rvert -1}{2}$ . Therefore $-\epsilon \gt \frac{1-\lvert L \rvert}{2}$. From which it follows:
$$ \lvert L \rvert - \epsilon \gt \frac{2 \lvert L \rvert +1 - \lvert L \rvert}{2}=\frac{\lvert L \rvert +1}{2}\gt 1$$
By our lemma, we then have:
$$\lvert L\rvert - \epsilon  \gt 1 \geq \lvert \sin(x) \rvert$$.
Rearranging:
$$ \lvert L \rvert \gt 1 + \epsilon \geq \lvert \sin(x) \rvert + \epsilon$$
$$ \lvert L \rvert - \lvert \sin(x) \rvert \gt \epsilon$$
Invoking one of the triangle inequality variants, we finally have:
$$ \lvert \sin(x) - L \rvert = \lvert L - \sin(x) \rvert \geq \lvert L \rvert - \lvert \sin(x) \rvert \gt \epsilon$$
Note that $\forall x \in \mathbb R$, the above statement is true. Therefore, any $N \in \mathbb R$ can be chosen and one can always find an $x \gt N$, for which the desired criteria is met.

Case 2: $\lvert L \rvert \leq 1$
Consider an $\epsilon \lt \frac{1}{2}$. For example, $\epsilon = \frac{1}{4}$
Here is where I am basically stumped in the absence of knowing anything about continuous functions. My hunch is that because $\lvert L \rvert \leq 1$, I can effectively assert that $\exists c \in \mathbb R \big [ \sin(c) = L \big ]$.
Now, invoking the same triangle inequality variant as before, we know that:
$$ \lvert \sin(c) - \sin(x) \rvert \geq \lvert \sin(c) \rvert - \lvert \sin(x) \rvert  $$
So if I can show that: $\exists x \in \mathbb R \big [\lvert \sin(c) \rvert - \lvert \sin(x) \rvert \gt \frac{1}{4} \big ]$, we are good to go. Once again, I assume that I need some notion of continuous functions in order to prove this.
More generally, in order to apply this argument to any $N \in \mathbb R$, I would like to prove the slightly more refined:
$$ \exists x \in [c - \pi, c + \pi] \text{ s.t. } \Big ( \lvert \sin(c) \rvert - \lvert \sin(x) \rvert \gt \frac{1}{4} \Big )$$
(noting the $c$ value that produces $\sin(c)=L$ can be replaced by $c+2\pi n$ for any any integer $n$).
If there is a method to circumvent continuous functions, I would greatly appreciate the insight!
Thank you!
 A: No knowledge of continuity is needed for this result. Suppose towards contradiction that $\lim_{x \to \infty} \sin(x) = L$. Then let $\epsilon = \frac{1}{4}$ and let $N$ be such that $x > N$ implies $|\sin(x) - L| < \frac{1}{4}.$
Choose integers $k, m > N$. Then clearly $2πk > N$ and $2πm + \frac{π}{2} > N$. But observe that:
\begin{align*}
|\sin(2πk) - L| &= |0 - L| = L < \frac1{4} \\
|\sin(2πm + \frac{π}{2}) - L| &= |1 - L| < \frac{1}4 \\
\end{align*}
Which is impossible, since $|1 - L| < \frac{1}{4} \implies L > \frac{3}{4}.$
A: 
i.e. I wanted to write a proof that:
$$\forall L \in \mathbb R \Big [\exists \epsilon \gt 0 \text{ s.t. } \forall N \in \mathbb R \ \exists x \in \mathbb R \text{ s.t. }\big (x \gt N \land \lvert \sin(x) - L \rvert \geq \epsilon \big) \Big]$$

We can do this using $\varepsilon = \frac{1}{2}$.
Case 1: $L > 0$
Here for any $N$ we can always find $x > N$ of the form $x = \frac{3\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
For such $x$ we have
$$\sin x = -1.$$
Since $L > 0$ we have
\begin{align}
|\sin x - L|&= |-1 - L| \\
& = 1 + L \\
& > \frac{1}{2}
\end{align}
Case 2 $L  < 0$
This case is nearly identical to the case for $L > 0$.
For any $N$ we can always find $x > N$ of the form $x = \frac{\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
For such $x$ we have
$$\sin x = 1.$$
Since $L < 0$ we have
\begin{align}
|\sin x - L|&= |1 - L| \\
& = 1 + |L| \\
& > \frac{1}{2}
\end{align}
Case 3 $L  = 0$
This case can be dealt with immediately by re-using either of the $x$ from the previous two cases:
If $x = \frac{\pi}{2} + 2k\pi$, then
\begin{align}
|\sin x - L|&= |1 - 0| \\
& = 1 \\
& > \frac{1}{2}
\end{align}
Of course, this all depends on knowing the behavior of $\sin$: that $\sin$ wavers between $1$ and $-1$ "forever" and so on. Given this behavior, one would expect intuitively that the limit doesn't exist, i.e. $\sin x$ wouldn't "settle" upon some single value for large $x$.
Later in the book Spivak defines the trigonometric functions formally in a way that conforms with the geometric/unit circle versions with which you are probably already familiar. Continuity plays a big part in these later definitions.
Here though, the main thing is just the periodic nature of $\sin$.
Finally, you may be somewhat leery of statements like:

...for any $N$ we can always find $x > N$ of the form $x = \frac{3\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.

This has at its heart the idea that given any real number $N$ we can find a natural number $n$ with $n > N$. This fact is employed in a few places early on in the text, but isn't formally proven until Chapter 8 (3rd Ed.) "Least Upper Bounds".
Spivak cops to this "cheating" at the end of that chapter.
(As an aside, an earlier problem in Chapter 5, 5-12 gives a very useful result:
If $f(x) \leq g(x)$ for all $x$ (or even just for all $x$ in a neighborhood of $a$) then $\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x)$, provided these limits exist.
This can be shown to hold if we replace $a$ with $\infty$.
This, combined with your lemma that $|\sin x| \leq 1$ can be used to rule out your Case 1, i.e. since $-1 \leq \sin x \leq 1$ we must also have $-1 \leq \lim_{x \to \infty}\sin x \leq 1$, if that limit exists.)
A: Note that since $\sin(x)$ is periodic for every real $N$  there exists a real $x>N$ and $y>N$ such that $\sin(x)=1$ and $\sin(y)=-1$  From here it’s easy to see that this implies if the limit exists then it must be $1$ since if it’s any other value $m$ then you can let $\epsilon$ be less than $|1-m|$ to obtain a contradiction . But the above argument also works for $-1$ a contradiction hence the limit does not exist .
