2018-08-12
展开全部
lim[x→0+] (1-e^√x)/√x = lim[x→0+] (√x)'(-e^√x)/(√x)' = lim[x→0+] (-e^√x) = -1 lim[x→0+] ln(1+√x)/√x = lim[x→0+] (√x)'/(1+√x)/(√x)' = lim[x→0+] 1/(1+√x) = 1 lim[x→0+] [√(1+√x)-1]/√x = lim[x→0+] (√x)'/[2√(1+√x)]/(√x)' = lim[x→0+] 1/[2√(1+√x)] = lim[x→0+] 1/[2√(1+√x)] = 1/2 lim[x→0+] (1-cos√x)/√x = lim[x→0+] (√x)'sin√x)/(√x)' = lim[x→0+] sin√x = 0 根据等价无穷小的定义,选B
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