# 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!

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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. –  Jonas Meyer Mar 15 '11 at 16:26
See mathkb.com/Uwe/Forum.aspx/math/16396/… (you'll probably find at least one satisfying answer there). –  Shai Covo Mar 15 '11 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? –  Uticensis Mar 16 '11 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. –  Arturo Magidin Mar 16 '11 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 $|e^{in}|=1$.

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

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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 '11 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 '11 at 17:28
@Nir: That is just a double angle formula. –  Jonas Teuwen Mar 15 '11 at 17:46
@Didier: Err... I will fix that. Well, anyway, I see that Henry added what I mean. –  Jonas Teuwen Mar 15 '11 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 $L$.

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