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# The limit of $$(\sin n)$$ 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 $$x\to +\infty$$.

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

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

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

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

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

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

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

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Note by Haroun Meghaichi
2 years, 6 months ago

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Obviously because there is not a set definition for the limit any oscillating series/function. · 2 years, 6 months ago

May I ask what math class you are in. · 2 years, 6 months ago

Haha, I'm in Algebra 1. I'm SUPPOSED to be in pre-calc or above, but my school-system is stupid. :D · 2 years, 6 months ago

Hi Haroun! Nice solution. One thing though, you are assuming that the limit of $$sin(n)$$ exists, for that the sequence $$sin(n)$$ needs to be monotonic and bounded. Obviously $$sin(n)$$ is bounded but what about monotonicity? It behaves differently( i.e. increasing/decreasing) in different intervals. · 2 years, 3 months ago

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

If a sequence is bounded and monotonic then it converges (this an implication not an equivalence). · 2 years, 3 months ago

A much easier way to see that $$sin(n)$$ can't possibly have a limit is to note that since $$sin(x)$$ 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.) · 2 years, 3 months ago

How you got sin2n tends to l ?? · 2 years, 6 months ago

If a sequence $$a_n$$ converge to some real $$l$$ then all of it subsequences converge to $$l$$ this follows from the limit definition. · 2 years, 6 months ago