How to calculate this limit?#lim_(n->oo)prod_(k=1)^ncos(2^(k-1)x);x!=kpi#

#sin(2^kx)=2cos(2^(k-1)x)sin(2^(k-1)x)#

or

#f_k = 2 cos(2^(k-1)x)f_(k-1)#
#f_(k-1) = 2 cos(2^(k-2)x)f_(k-2)#
#cdots#
#f_1 = 2 cos(x)f_0#

then

#f_k = (2^k prod_(j=0)^k cos(2^jx)) f_0#

then

# prod_(j=0)^n cos(2^jx)=1/2^n(f_n)/(f_0)=1/2^n(sin(2^n x)/sinx)#

we can then conclude that for #x ne k pi#

#lim_(n->oo)prod_(j=0)^n cos(2^jx)=0#

because if #x ne k pi# with #k in ZZ# we have

#abs(sin(2^n x)/sinx) le M#

so there exists #n# such that #M/2^n < epsilon# for arbitrarily small #epsilon#.