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.

If you liked it, share it. If you have question just write them.

## Comments

Sort by:

TopNewestObviously because there is not a set definition for the limit any oscillating series/function. – Finn Hulse · 3 years, 1 month ago

Log in to reply

– Kahsay Merkeb · 3 years, 1 month ago

May I ask what math class you are in.Log in to reply

– Finn Hulse · 3 years, 1 month ago

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

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. – Jit Ganguly · 2 years, 10 months ago

Log in to reply

If a sequence is bounded and monotonic then it converges (this an implication not an equivalence). – Haroun Meghaichi · 2 years, 10 months ago

Log in to reply

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.) – Austin Stromme · 2 years, 10 months ago

Log in to reply

How you got sin2n tends to l ?? – Vivek Sedani · 3 years, 1 month ago

Log in to reply

– Haroun Meghaichi · 3 years, 1 month 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.Log in to reply

sin(n)cos(n). and hence sin2n= 2l(sqrt(1-l^2))? – Vivek Sedani · 3 years agoLog in to reply