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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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5 Answers 5

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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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  • $\begingroup$ hey jonas, can you please explain me the transformation fron sin to cos and what to do afterwards a little bit more? $\endgroup$
    – user6163
    Mar 15, 2011 at 17:26
  • $\begingroup$ 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. $\endgroup$
    – Did
    Mar 15, 2011 at 17:28
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    $\begingroup$ @Nir: That is just a double angle formula. $\endgroup$
    – JT_NL
    Mar 15, 2011 at 17:46
  • $\begingroup$ @Didier: Err... I will fix that. Well, anyway, I see that Henry added what I mean. $\endgroup$
    – JT_NL
    Mar 15, 2011 at 17:47
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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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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – andrewBee
    Nov 1, 2021 at 17:38