# Prove that the limit of $\sin n$ as $n \rightarrow \infty$ does not exist [duplicate]

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!

• Follows from stronger result at math.stackexchange.com/questions/9319/…; Related: math.stackexchange.com/questions/22047/sinn-is-not-u-d-mod-1; 2 more versions of first link: math.stackexchange.com/questions/4764/…, math.stackexchange.com/questions/7252/… Mar 15, 2011 at 16:23
• This question is asking for something weaker than the density asked for in previous problems, and correspondingly is easier to answer, and therefore I am unsure whether it should be considered a duplicate. Mar 15, 2011 at 16:26
• See mathkb.com/Uwe/Forum.aspx/math/16396/… (you'll probably find at least one satisfying answer there). Mar 15, 2011 at 17:44
• @Arturo Magidin There was something obviously wrong with my notation the first time I edited this question. For future reference, what was it, so I can be more observant in my own problem-solving? Mar 16, 2011 at 12:08
• @Billare: $\{a_n\}$ is interpreted as a sequence (or a family of terms); $\{a_n\}=\sin n$ just looks wrong: $\sin n$ is itself not a sequence or a set of values. Mar 16, 2011 at 15:46

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 $|e^{in}|=1$.

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 $\sin(2(n + 1) - 2n)$.

• hey jonas, can you please explain me the transformation fron sin to cos and what to do afterwards a little bit more?
– user6163
Mar 15, 2011 at 17:26
• There exists infinitely many integers $n$ such that $\sin(n)$, $\sin(n+1)$ and $\sin(n+2)$ are all in the interval $[\frac12,1]$, so I fail to see the conclusion of the argument in your last paragraph.
– Did
Mar 15, 2011 at 17:28
• @Nir: That is just a double angle formula. Mar 15, 2011 at 17:46
• @Didier: Err... I will fix that. Well, anyway, I see that Henry added what I mean. Mar 15, 2011 at 17:47

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 $L$.

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.

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.

• The addition formula is wrong here. The idea is correct, though. There is a subtlety about why the limit of cosine is given by the Pythagorean theorem since there could be problems with the \pm sign when taking the square root. These issues can be avoided. Nov 1, 2021 at 17:38