因cos x /2cosx/4…cosx/2^n=[cosx/2*cosx/4*.*2sinx/2^n*cosx/2^n]/(2sinx/2^n)=[cosx/2*cosx/4*...*sinx/2^(n-1)]/(2sinx/2^n)=(cosx/2sinx/2)/[2^(n-1)*sin(x/2^n]=sinx/[2^n*sin(x/2^n)]所以lim (n趋近正无穷) c...