Prove that the limit of $\sin n$ as $n \rightarrow \infty$ does not exist Using only the delta definition of a limit, how can we prove that the sequence $\{a_n\}$, where $a_n = \sin n$, as $n$ tends to infinity does not have a limit?
Thanks!
 A: The following are true, based on standard trigonometric identities and $\sin(1) \approx 0.84147$ and $\sin(3) \approx 0.14112$:
$$\begin{align}
\textrm{if } \sin(n) \le -0.4, & \textrm{ then } 0 < \sin(n+3) ; \\
\textrm{if } -0.4 \le \sin(n) \le 0.4, & \textrm{ then } \sin(n+1) < -0.4 \textrm{ or } 0.4 < \sin(n+1) ; \\
\textrm{if } 0.4 \le \sin(n),& \textrm{ then } \sin(n+3) < 0;
\end{align}$$
so there is no value $L$ where for any positive $\varepsilon < 0.2$ you have all of $\sin(n), \sin(n+1), \sin(n+3)$ and $\sin(n+4)$ within $\varepsilon$ of $L$.     
A: In any interval of the form $ [k\pi +\frac{\pi}{3},k\pi +\frac{2\pi}{3}]$, where $k $ is any natural number, there is at least a natural number $n_{k}$. The reason is that any such interval has length $\pi/3$ that is greater than 1. Since those intervals are mutually disjoint then the sequence $\{n_{k}\}$ is a sub-sequence of the sequence $\{n\}$ and, obviously, $|\sin n_k|\geq \frac{\sqrt{3}}{2}$. In a similar way, considering the intervals of the form $[k\pi -\frac{\pi}{6},k\pi +\frac{\pi}{6}]$, we can construct another sub-sequence $\{m_k\},$  of the sequence $\{n\},$ such that $|\sin m_k|\leq \frac{1}{2}$. Assume that $\lim_{n\to \infty}\sin{n}$ exists and is the number $l$. Using both sub-sequences defined above, we obtain that $|l|\geq \frac{\sqrt{3}}{2}$ and $|l|\leq \frac{1}{2}$, and this is a contradiction. 
A: No need for $\epsilon$ actually.
If $\sin(n) \rightarrow l$, then $\sin(n+1)$ also, and $\sin(n+1)=\sin(n)\cos(1)+\sin(1)\cos(n)$.
Since both $\sin(n)$ and $\sin(n+1)$ have limit $l$ and $\sin(1) \neq 0$, $\cos(n) \rightarrow \frac{l(1-\cos(1))}{\sin(1)}$, and so $e^{in}=\cos(n)+i \sin(n)$ has a limit.
But $e^{i(n+1)}$ must then have the same limit (call it $x$), which implies $x=e^{i} x$, and since $e^{i} \neq 1$, $x$ has to be zero, which is a contradiction with the fact that $|e^{in}|=1$.
A: Assume $\lim \sin(n) = l$. Then so is $\lim \sin(2n) = l$. So $\lim \cos(2n) = 1 - 2l^2$, but so does the limit of $\cos(2(n + 1))$. Now apply the sum-formula to $\sin(2(n + 1) - 2n)$.
A: Let if possible $\sin n\rightarrow x$. Then $\sin k=\sin(n+k-n)=\sin(n+k)\cos k-\cos(n+k)\sin n\rightarrow x(\cos k-\sqrt{1-x^2})$  for each positive integer k. Now as $k\rightarrow \infty$ implies that $x=x.0=0$ which shows that $\sin k=0$ for all k...a contradiction.
