求一数学题：(cosx/2)(cosx/2^2)(cosx/2^3).......(cosx/2^n)当n趋近于无穷时的极限，注意x为常数

(cosx/2)(cosx/2^2)(cosx/2^3).......(cosx/2^n)
=(cosx/2)(cosx/2^2)(cosx/2^3).......(cosx/2^n)*(sinx/2^n)/(sinx/2^n)
=(1/2)^1*(cosx/2)(cosx/2^2)(cosx/2^3).......(cosx/2^(n-1))sin(x/2^(n-1)/sin(x/2^n)
=……
=(1/2)^(n-1)*cosx/2*sinx/2 / sin(x/2^n)
=sinx/(2^n*sin(x/2^n))

已知对于在一定范围内的sinx,有
x/(1+x)< sinx < x
当n充分大时
x/2^n在此范围内，
有
x/(2^n+x)<sin(x/2^n) < x/2^n
=>
x/(1 + x/2^n)< 2^n*sin(x/2^n) <x
所以有夹逼定理，n趋于无穷时，
sinx/(2^n*sin(x/2^n)) = sinx/x
=>
x不为0时
(cosx/2)(cosx/2^2)(cosx/2^3).......(cosx/2^n)=sinx/x
若x=0
(cosx/2)(cosx/2^2)(cosx/2^3).......(cosx/2^n)=1