#lim_(n->oo)(Sin n+cos n)^n# Find the value? Thanks

2 Answers
Mar 4, 2018

See below.

Explanation:

#sinn+cosn = sqrt2 sin(n+pi/4)#

#(sinn+cosn)^n = 2^(n/2) sin^n(n+pi/4)#

but #(n+ pi/4)/(2pi)# is irrational so #a_n = sin(n+pi/4)# is not periodic and can take any value in #[-1,1]# and then concluding

#lim_(n->oo)(sinn+cosn)^n # does not exists.

Mar 4, 2018

The limit doesn't exist.

Explanation:

Since #sin# and #cos# are periodic functions, the limits as #n# approaches infinity of #sin n# and #cos n# #color(red)("never")# converge to a value.