Non-existing Limit of $\sin x$ How do I prove from definition of limit that $\lim_{x \to \infty}\sin x$ is non-existant? I tried to negate said definition:
$$\lnot ((\exists L)(\forall\epsilon)(\exists \delta):(\forall x)(|x|\gt \delta)\Rightarrow(|\sin x-L|\lt\epsilon)) = ((\forall L)(\exists\epsilon)(\forall \delta):(\exists x)(|x|\gt \delta)\land(|\sin x-L|\ge\epsilon))$$ , but I am not sure I negate it right, and I have little idea how to prove that statement.

Your formal negation seems fine, but it probably requires some verbal elaboration in order for the proof idea to come naturally.
Saying that such a limit exists means that there is some value $L$ such that by going suitably far along the real line, forces the values of $\sin(x)$ to become close to that $L$.  Intuitively, it is clear that $\sin(x)$ does not have a horizontal asymptote, and that its values perpetually bounce between $-1 $ and $1$.
I'm going to leave formalizing all that to you, but I will note that you can take advantage of periodicity to note that $\sin(2\pi k + \pi/2) = 1, \sin(2\pi k + 3\pi/2) = -1$ for all integers $k$, and that both sets $\{2\pi k + \pi/2\}, \{2\pi k + 3\pi/2\}$ are unbounded.
