1个回答
展开全部
lim ln[1+e^(2/x)]/ln[1+e^(1/x)]
x→0+
=lim {e^(2/x)·(-2/x²)/[1+e^(2/x)]}/{e^(1/x)·(-1/x²)/[1+e^(1/x)]}
x→0+
=lim 2[e^(1/x)+e^(2/x)]/[1+e^(2/x)]
x→0+
=lim 2[e^(1/x)·(-1/x²)+e^(2/x)·(-2/x²)]/[e^(2/x)·(-2/x²)]
x→0+
=lim [1/e^(1/x)+2]
x→0+
=0+2
=2
x→0+
=lim {e^(2/x)·(-2/x²)/[1+e^(2/x)]}/{e^(1/x)·(-1/x²)/[1+e^(1/x)]}
x→0+
=lim 2[e^(1/x)+e^(2/x)]/[1+e^(2/x)]
x→0+
=lim 2[e^(1/x)·(-1/x²)+e^(2/x)·(-2/x²)]/[e^(2/x)·(-2/x²)]
x→0+
=lim [1/e^(1/x)+2]
x→0+
=0+2
=2
本回答被提问者采纳
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