求第九题详细解答过程。。。
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先观察函数本身，分子分母同乘以(2^n)*sin[x/2^n]
数列={[cos(x/2)cos(x/4)...cos(x/2^n)]*(2^n)*sin(x/2^n)}/[(2^n)*sin(x/2^n)]
={[cos(x/2)cos(x/4)...cos(x/2^(n-1)]*[2^(n-1)]*sin[x/2^(n-1)]}/[(2^n)*sin(x/2^n)]
={[cos(x/2)cos(x/4)...cos(x/2^(n-2)]*[2^(n-2)]*sin[x/2^(n-2)]}/[(2^n)*sin(x/2^n)]
=......
=[cos(x/2)*2*sin(x/2)]/[(2^n)*sin(x/2^n)]
=sinx/[(2^n)*sin(x/2^n)]
利用等价无穷小代换
原极限=lim(n->∞)sinx/[(2^n)*(x/2^n)]
=sinx/x
数列={[cos(x/2)cos(x/4)...cos(x/2^n)]*(2^n)*sin(x/2^n)}/[(2^n)*sin(x/2^n)]
={[cos(x/2)cos(x/4)...cos(x/2^(n-1)]*[2^(n-1)]*sin[x/2^(n-1)]}/[(2^n)*sin(x/2^n)]
={[cos(x/2)cos(x/4)...cos(x/2^(n-2)]*[2^(n-2)]*sin[x/2^(n-2)]}/[(2^n)*sin(x/2^n)]
=......
=[cos(x/2)*2*sin(x/2)]/[(2^n)*sin(x/2^n)]
=sinx/[(2^n)*sin(x/2^n)]
利用等价无穷小代换
原极限=lim(n->∞)sinx/[(2^n)*(x/2^n)]
=sinx/x
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