毛笔巫师
幼苗
共回答了17个问题采纳率:88.2% 举报
题目是:lim(x-0) (secx-1)/(x^2)吗?是的话,
lim(x-0) (secx-1)/(x^2)=lim(x-0) (1/cosx-1)/(x^2)=lim(x-0) [(1-cosx)/cosx]/(x^2)
后面用等价无穷小作替换:当(x-0)时,1-cosx与x^2/2等价无穷小
因此:lim(x-0) [(1-cosx)/cosx]/(x^2)=lim(x-0) [(x^2/2)/cosx]/(x^2)=lim(x-0)1/2cosx=1/2
1年前
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