Showing that the limit of $\sin (n) $ as $n $ approaches infinity doesn't exist In order to do that I want to find two subsequences $a_n$ and $b_n$ of $\sin (n) $ such that $\lim_{n \to +\infty} a_n $ and $\lim_{n \to +\infty} b_n$ are not the same number. 
Hints?
 A: Well if it approached some limit $L$, then $\sin^2 n$ approaches $L^2$ and $\cos^2 n$ approaches $1 - L^2$. For any integer $a$, we use the identity
$$\sin(n + a) = \sin n \cos a + \cos n \sin a$$
This gives 
$$(\sin(n + a) - \sin n \cos a)^2 = \cos^2 n \sin^2 a$$
Now take the limit as $n$ goes to infinity, and you get
$$(L - L \cos a)^2 = (1 - L^2) \sin^2 a$$
Equivalently,
$${L^2 \over 1 - L^2} = \bigg({\sin a  \over 1 - \cos a}\bigg)^2$$
(Take reciprocals of this if $L^2 = 1$).
However, ${\displaystyle \bigg({\sin a  \over 1 - \cos a}\bigg)^2}$ is different for $a = 1$ and $2$ for example. Hence the limit cannot exist.
A: Each unit interval center on $\pi/2 + 2k\pi$,
$$ A_k = \left[ \frac{\pi}{2} + 2k\pi - \frac 12, \frac{\pi}{2} + 2k\pi + \frac 12\right] \quad\text{ for integers } k > 0$$ 
contains a unique positive integer. Call the sequence of such integers $a_k$.
If $\lim_{n\to\infty} \sin(n)$ exists then $\lim_{k\to\infty} \sin(a_k)$ also exists and is equal to that limit, call it $L$. Note that $L$ is positive as
$$L \geq \min_{k,x} \left\{ \sin x \ : \ x \in A_k  \right\} = \min_x \left\{ \sin x \ : \ x \in \left[\frac \pi 2- \frac 12,\frac \pi 2+ \frac 12 \right]  \right\} = \sin\left(\frac \pi 2- \frac 12\right) > 0$$
Now if we construct another sequence for integers in unit intervals centered on $3\pi/2 + 2k\pi$ we can bound $L < 0$. Contradiction and therefore no such $L$ exists.
A: A limit such as yours exists if corresponding $\limsup$ and $\liminf$ exist and are equal. For $\sin(x)$ as $x$ goes to infinity, $$ \limsup_{x \to \infty} \sin(x)=1$$ while $$ \liminf_{x \to \infty}\sin(x)=-1 .$$
Since these two limits are not equal, the original limit does not exist.  When something is obvious the trick is often to return to definitions.
A: Hint: 
1) $a_n=n\pi$, $n\in\mathbb{Z}$
2) $b_n=(2n+1)\frac{\pi}{2}$, $n\in\mathbb{Z}$
What can you say about the value of $\sin$ evaluated at each term of the sequences $a_n,b_n$?
