How to prove that $\lim_{n\to\infty}{|\sin n|}$ doesn't exist

Question

How to prove that $\lim_{n\to\infty}{|\sin n|}$ doesn't exist. $n\in\mathbb N$

Any hint?

Answer 1

What do we mean a limit exist? We should have a number $l\in R$ (or $R\cup\{\infty\}\cup\{-\infty\})$ such that $$\lim_{n\to\infty}| |\sin n|-l|=0\,\,\,\, \,\,\,\,\,\,\,\,\,\,(1)$$ That is, for any subsequence of $(n)$ the equation $(1)$ should hole.

Now, let us choose $n_k\in[2k\pi-\pi/6,2k\pi +\pi/6]$ be an integer, then we have $$ \limsup_{k\to\infty}|\sin n_k|\leq \frac{1}{2} $$ However, if we choose $n_k\in [2k\pi+\pi/4,2k/pi+3\pi/4]$ be an integer (choose when you can), then $$ \liminf_{k\to\infty}|\sin n_k|\geq \sqrt{2}/2 $$

Hence the limit does not exist.

Note that we are keep choose an subsequence of sequence $(n)$. The subsequence, indexed by $k$, does not need exists for each $k$,

Answer 2

For $x \in \left[2k\pi + \frac{\pi}{6} , 2k\pi + \frac{5\pi}{6}\right]$ for integer $k$, you have $\sin(x) \ge \frac12$.

For $x \in \left[2k\pi + \frac{7\pi}{6} , 2k\pi + \frac{11\pi}{6}\right]$ for integer $k$, you have $\sin(x) \le -\frac12$.

So for every $7(\gt 2\pi)$ consecutive integers you have at least two with $\sin(n) \ge \frac{1}{2}$ and at least two with $\sin(n) \le -\frac{1}{2}$, since $\frac{5\pi}{6}-\frac{\pi}{6}=\frac{11\pi}{6}-\frac{7\pi}{6}\gt 2$.

So $\sin(n)$ does not converge.

Answer 3

Assume there is a limit $L$. This means given any $\epsilon > 0$, eventually we have $$\vert \sin(n) - L \vert < \epsilon \implies \sin(n) - \epsilon < L < \sin(n) + \epsilon$$

To be specific, let us pick $\epsilon = 1/4$.

Now pick $n \in (2m \pi - \pi/2,2m \pi -\pi/6)$. This is possible since the interval length is $\pi/3>1$. We have $\sin(n) < -1/2$. There exists infinitely many $n$'s and hence $L < -1/4$.

Now pick $n \in (2m \pi + \pi/6,2m \pi +\pi/2)$. This is possible since the interval length is $\pi/3>1$. We have $\sin(n) > 1/2$. There exists infinitely many $n$'s and hence $L>1/4$.

Hence, no such $L$ exists.