Limit of [1 + sin(nπ/3)cos(nπ/5)]/√n using sandwich theorem

[CathyC]

1. Use the sandwich theorem to compute the limit as n goes to infinity of the sequence with the following nth elements:

a(n) = [1 + sin(n*pi/3)cos(n*pi/5) ] / [n^0.5]

I would really appreciate some help with this one guys. If you could please go slow with the answer as my trig is pretty shaky. Thanks for all your help! :)

Cathy

 

[Dick]

You don't really have to use a lot of trig. -1<=sin(x)<=1 and the same for cos(x). No matter what x is. Suggest an upper bound for the value of the numerator.