The limit of (sinn)(\sin n) does not exist.

Hello there,

Maybe the first thing that a calculus student when studying the sine and cosine function is that these are the first that he knows and do not have a limit as x+x\to +\infty.

But how about the sequence (sinn)(\sin n) and (cosn)(\cos n) ? do they have a limit ?

A method to prove that (sinn)(\sin n) does not have a limit is as follow :

  • Assume that it has a limit then it should be a real number ll since sin\sin is bounded.

  • Since sinnl\sin n\to l, we get sin(2n)l\sin(2n) \to l and : cos(2n)12l2\cos(2n) \to 1-2l^2.

  • we have : sin2=sin(2n+22n)=sin(2n+2)cos(2n)sin(2n)cos(2n+2).\sin 2 = \sin(2n+2-2n) =\sin(2n+2)\cos(2n) -\sin(2n)\cos(2n+2).

But the limit of the right side is : l(12l2)l(12l2)=0. l(1-2l^2) -l(1-2l^2) =0.

Which means that sin2=0\sin 2=0 which is absurd.

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Note by Haroun Meghaichi
7 years, 2 months ago

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Hi Haroun! Nice solution. One thing though, you are assuming that the limit of sin(n)sin(n) exists, for that the sequence sin(n)sin(n) needs to be monotonic and bounded. Obviously sin(n)sin(n) is bounded but what about monotonicity? It behaves differently( i.e. increasing/decreasing) in different intervals.

Jit Ganguly - 6 years, 11 months ago

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A convergent sequence does not have to be monotonic, for example (1)nn\frac{(-1)^n}{n}.

If a sequence is bounded and monotonic then it converges (this an implication not an equivalence).

Haroun Meghaichi - 6 years, 11 months ago

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Obviously because there is not a set definition for the limit any oscillating series/function.

Finn Hulse - 7 years, 2 months ago

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May I ask what math class you are in.

Kahsay Merkeb - 7 years, 2 months ago

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Haha, I'm in Algebra 1. I'm SUPPOSED to be in pre-calc or above, but my school-system is stupid. :D

Finn Hulse - 7 years, 2 months ago

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In any interval of the form [kπ+π3,kπ+2π3] [k\pi +\frac{\pi}{3},k\pi +\frac{2\pi}{3}], where kk is any natural number, there is at least a natural number nkn_{k}. The reason is that any such interval has length π/3\pi/3 that is greater than 1. Since those intervals are mutually disjoint then the sequence {nk}\{n_{k}\} is a sub-sequence of the sequence {n}\{n\} and, obviously, sinnk32|\sin n_k|\geq \frac{\sqrt{3}}{2}. In a similar way, considering the intervals of the form [kππ6,kπ+π6][k\pi -\frac{\pi}{6},k\pi +\frac{\pi}{6}], we can construct another sub-sequence {mk},\{m_k\}, of the sequence {n},\{n\}, such that sinmk12.|\sin m_k|\leq \frac{1}{2}. Assume that limnsinn\lim_{n\to \infty}\sin{n} exists and is the number l.l. Using both sub-sequences defined above, we obtain that l32|l|\geq \frac{\sqrt{3}}{2} and l12|l|\leq \frac{1}{2}, and this is a contradiction.

Arturo Presa - 3 years, 3 months ago

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How you got sin2n tends to l ??

vivek sedani - 7 years, 2 months ago

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If a sequence ana_n converge to some real ll then all of it subsequences converge to ll this follows from the limit definition.

Haroun Meghaichi - 7 years, 2 months ago

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but how are you taking sin2n = l because as you are taking cos2n = 1 -2* l^2. can't we take sin2n=2sin(n)cos(n). and hence sin2n= 2l(sqrt(1-l^2))?

vivek sedani - 7 years, 1 month ago

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A much easier way to see that sin(n)sin(n) can't possibly have a limit is to note that since sin(x) sin(x) doesn't have a limit and is continuous, it can't possibly be that the limit at integral values exists. (I'll leave it to the interested reader to prove this rigorously.)

Austin Stromme - 6 years, 11 months ago

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