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.