
Re: Sequence limit
Posted:
Oct 3, 2013 4:49 PM


On Thursday, October 3, 2013 10:26:13 PM UTC+2, quasi wrote: > quasi wrote: > > >konyberg wrote: > > >>Bart Goddard wrote: > > >>> > > >>> This question from a colleague: > > >>> > > >>> What is lim_{n > oo} sin n^(1/n) > > >>> > > >>> where n runs through the positive integers. > > >>> > > >>> Calculus techniques imply the answer is 1. > > >>> But the same techniques imply the answer is 1 > > >>> if n is changed to x, a real variable, and that > > >>> is not the case, since sin x =0 infinitely often. > > >>> > > >>> Anyone wrestled with the subtlies of this problem? > > >>> > > >>> E.g., can you construct a subsequence n_k such > > >>> that sin (n_k) goes to zero so fast that the > > >>> exponent can't pull it up to 1? > > >> > > >>In general a^0 = 1. lim (n goes inf) 1/n = 0. Then the value > > >>of sin(n) doesn't change that a^0 = 1. > > > > > >Your logic is flawed. > > > > > >Let f(n) = 1/(2^n). > > > > > >Then f(n)^(1/n) = 1/2 for all nonzero values of n, hence the > > >limit, as n approaches infinity, of f(n)^(1/n) is 1/2, not 1. > > > > > >Are there infinitely many positive integers n such that > > > > > > sin(n)^(1/n) < 1/(2^n) > > > > I meant: > > > > Are there infinitely many positive integers n such that > > > > sin(n) < 1/(2^n) > > > > >?? > > > > > >If so, then the limit of the sequence > > > > > > sin(n)^(1/n), n = 1,2,3, ... > > > > > >does not exist. In particular, it would not be equal to 1. > > > > > >In fact, the original question can be recast as: > > > > > >Does there exist a real number c with 0 < c < 1 such that > > >the inequality > > > > > > sin(n) < c^n > > > > > >holds for infinitely many positive integers n? > > > > quasi
Hi.
Yes I was a bit hasty here. But sin(n) is limited from 1 to +1 (your function isn't limited), and the limit of 1/n is 0. I would think that the limit is 1. The debate will still be what 0^0 is equal to :)
KON

