展开全部
method 1:
lim(n->∞) ( cos(1/n) )^(n^2)
=lim(n->∞) ( 1- (1/2)(1/n)^2 )^(n^2)
=e^(-1/2)
method 2:
consider
lim(x->∞) ( cos(1/x) )^(x^2)
y=1/x
=lim(y->0+) ( cosy )^(1/y^2)
=lim(y->0+) e^[ ln(cosy )/y^2]
=lim(y->0+) e^[ ln(1 -(1/2)y^2 )/y^2]
=lim(y->0+) e^[ -(1/2)y^2 /y^2]
=e^(-1/2)
=>
lim(n->∞) ( cos(1/n) )^(n^2) = e^(-1/2)
method 3:
L = lim(x->∞) ( cos(1/x) )^(x^2)
lnL
=lim(x->∞) x^2.ln[ cos(1/x) ]
=lim(x->∞) ln[ cos(1/x) ] /(1/x^2) (0/0 分子分母分别求导)
=lim(x->∞) (1/x^2).tan(1/x) /(-2/x^3)
=-(1/2) lim(x->∞) x.tan(1/x)
=-1/2
=>lim(x->∞) ( cos(1/x) )^(x^2) = e^(-1/2)
=>lim(n->∞) ( cos(1/n) )^(n^2) = e^(-1/2)
lim(n->∞) ( cos(1/n) )^(n^2)
=lim(n->∞) ( 1- (1/2)(1/n)^2 )^(n^2)
=e^(-1/2)
method 2:
consider
lim(x->∞) ( cos(1/x) )^(x^2)
y=1/x
=lim(y->0+) ( cosy )^(1/y^2)
=lim(y->0+) e^[ ln(cosy )/y^2]
=lim(y->0+) e^[ ln(1 -(1/2)y^2 )/y^2]
=lim(y->0+) e^[ -(1/2)y^2 /y^2]
=e^(-1/2)
=>
lim(n->∞) ( cos(1/n) )^(n^2) = e^(-1/2)
method 3:
L = lim(x->∞) ( cos(1/x) )^(x^2)
lnL
=lim(x->∞) x^2.ln[ cos(1/x) ]
=lim(x->∞) ln[ cos(1/x) ] /(1/x^2) (0/0 分子分母分别求导)
=lim(x->∞) (1/x^2).tan(1/x) /(-2/x^3)
=-(1/2) lim(x->∞) x.tan(1/x)
=-1/2
=>lim(x->∞) ( cos(1/x) )^(x^2) = e^(-1/2)
=>lim(n->∞) ( cos(1/n) )^(n^2) = e^(-1/2)
本回答被提问者和网友采纳
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
广告 您可能关注的内容 |