跪求两道数学题!一道极限一道函数!!!!!!!!!

2024-12-22 23:53:33
推荐回答(2个)
回答1:

1、cos(x/2)cos(x/4)...cos(x/2^n)
=2sin(x/2^n)cos(x/2)cos(x/4)...cos(x/2^n)/2sin(x/2^n)
=cos(x/2)cos(x/4)...cos(x/2^(n-1))sin(x/2^(n-1))/2sin(x/2^n)
=cos(x/2)cos(x/4)...cos(x/2^(n-2))sin(x/2^(n-2))/4sin(x/2^n)
=......
=cos(x/2)sin(x/2)/2^(n-1)sin(x/2^n)
=sinx/2^nsin(x/2^n)
所以原式=lim(n->∞) sinx/[2^nsin(x/2^n)]
=lim(n->∞) sinx/[2^n(x/2^n)]
=(sinx)/x
2、f(x)=(x-1+1)/(x-1)=1+1/(x-1)

f[f(x)]=1+1/[1+1/(x-1)-1]=x
f{f[f(x)]}=x/(x-1)

回答2:

f[f(x)]=1+1/[1+1/(x-1)-1]=x f{f[f(x)]}=x/(x-1)