Question

大学微积分课后题，求大神解答第8题

Answer 1

证明 ：
∵sinx=2cos(x/2)sin(x/2)=2^2cos(x/2)cos(x/2^2)sin(x/2^2)=...
=2^ncos(x/2)cos(2/2^2)...cos(x/2^n)sin(x/2^n)
∴cos(x/2)cos(2/2^2)...cos(x/2^n)=sinx/(2^nsin(x/2^n)
∵当n->∞时, sin(x/2^n) ~x/2^n 是等价无穷小
∴lim(n->∞)(cos(x/2)cos(2/2^2)...cos(x/2^n))
=lim(n->∞)(sinx/(2^nsin(x/2^n)))
=lim(n->∞)(sinx/(2^n*x/2^n))
=lim(n->∞)(sinx/x)
=sinx/x