Question

数列极限的问题

lim[(n2+1/n+1)-na-b]=0 求实数a和b
n2是n的平方

Answer 1

方法一
0=lim[(n^2+1)/(n+1)-na-b]
=lim[(1-a)n^2-(a+b)n+(1-b)]/(n+1) (通分)
首先(1-a)=0, 即a=1, 否则极限为无穷
于是极限为a+b，故a+b=0, b=-1

方法二
0=lim[(n^2+1)/(n+1)-na-b]
0=lim[(n^2+1)/(n+1)-na-b]/n
=lim[(n^2+1)/(n(n+1))-a-b/n]
=1-a-0
=1-a
故a=1且
0=lim[(n^2+1)/(n+1)-n-b]
=lim[(n-1)+2/(n+1)-n-b]
=lim[-1+2/(n+1)-b]
=-1-b
故b=-1
于是a=1，b=-1

Answer 2

详细解答,参见图解。点击放大，再点击再放大。