求第九题详细解答过程。。。
展开全部
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
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
