极限问题过程 10
展开全部
cos(xy) +lny -x = 1
x=0
1 +lny(0) -0 = 1
lny(0) = 0
y(0) =1
cos(xy) +lny -x = 1
两边求导
(-sin(xy)).( y + xy') + (1/y)y' - 1 =0
{ [xysin(xy) -1] /y } y' = -ysin(xy) -1
y'= -y.[ysin(xy) +1] /[xysin(xy) -1]
y'(0) = -[0 +1] /[0 -1] =1
consider
let
y=1/x
lim(x->∞) x[ f(2/x) -1]
=lim(y->0) [ f(2y) -1]/y (0/0分子分母分别求导)
=lim(y->0) 2f'(2y)
=2f'(0)
=2
=>
lim(n->∞) n[ f(2/n) -1] =2
ans : A
本回答被网友采纳
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