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.

We know that there is a sequence of points $x_n \rightarrow \infty $ with $ \sin (x_n) = (-1)^n $, eg take $x_n = \frac{2n+1}{2}\pi $.  If $\exists L$ such that $\sin (x) \rightarrow L$ as $x \rightarrow \infty$ we need to have that $\forall \epsilon > 0 ~ ~ \exists R \in \mathbb{R}$ such that $x>R \Rightarrow |\sin(x) - L | < \epsilon $.
A necessary condition for this would be that $\forall \epsilon > 0 ~ \exists N \in \mathbb{N}$ with $ n>N \Rightarrow |\sin(x_n)-L|<\epsilon$.
Well, observe that taking $\epsilon < 1 $ we find that for all guesses of $L$ and all $ n \in \mathbb{N}$ at least one of $| \sin(x_n) - L|$ and $|\sin(x_{n+1})-L|$ is greater than $\epsilon$: but if $L$ were a limit this would have to not be the case for $n$ sufficiently large.  Thus no such $L$ can exist.
I hope this isn't too wordy.  Your negation of the statement is correct, $\delta$ is usually suggestive of a small number but that's a matter of taste and doesn't affect the correctness of what you've written.
