求证lim(n→∞)(2n^2+1)/(3n^2+1)=2/3
1个回答
展开全部
对于任意ε>0
取N=max{1,1/(3根号ε)}
当n>N时有
|(2n^2+1)/(3n^2+1)-2/3|
=|[6n^2+3-6n^2-2]/[3(3n^2+1)]|
=1/[3(3n^2+1)]
<1/[3(3n^2)]
=1/(9n^2)
<1/(9/(9ε))
=ε
(因为3n^2+1>3n^2>0
所以1/(3n^2+1)<1/(3n^2))
所以，由极限定义
lim(n→∞)(2n^2+1)/(3n^2+1)=2/3
取N=max{1,1/(3根号ε)}
当n>N时有
|(2n^2+1)/(3n^2+1)-2/3|
=|[6n^2+3-6n^2-2]/[3(3n^2+1)]|
=1/[3(3n^2+1)]
<1/[3(3n^2)]
=1/(9n^2)
<1/(9/(9ε))
=ε
(因为3n^2+1>3n^2>0
所以1/(3n^2+1)<1/(3n^2))
所以，由极限定义
lim(n→∞)(2n^2+1)/(3n^2+1)=2/3
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
