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`.