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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.

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  • $\begingroup$ by $sinx$ did you mean $\sin(x)$? $\endgroup$ – ja72 Apr 14 '15 at 17:12
  • $\begingroup$ Yes, you have the right negation. $\endgroup$ – Tim Raczkowski Apr 14 '15 at 17:15
  • $\begingroup$ As you want it, the only way I can see (and I'm not sayin there could be some short cuts) is to do several cases...and it is going to be quite lengthy and boring. Of course, the easiest way is to show two sequences converging to infinity such that sine on each of them converges to something different. $\endgroup$ – Timbuc Apr 14 '15 at 17:15
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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.

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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.

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$\lim_{x\to\infty}\sin x$, if exists, it would be $\lim_{n\to\infty}\sin x_n$ for any sequence $x_n$ tending to infinity.

(https://en.wikipedia.org/wiki/Limit_of_a_function look for Heine's definition)

A classical result is that $\{\sin n : n\in\mathbb{N}\}$ is dense in $[-1,1]$, hence for $x_n=n$ the limit doesn't exist, so $\lim_{x\to\infty}\sin x$ doesn't exist.

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