若lim (an的平方+bn-5)/(2n+1)=1 为什么a=0否则极限就不存在
数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 求a b的值
高二的极限运算题 lim(2an+4bn)=8,lim(6an-bn)=1,求lim(3an+bn)的值 n趋向于无穷大
等差数列{an},{bn}的前n项和分别为An,Bn,切An/Bn=2n/3n+1,求lim(n→∞)an/bn
关于数列的极限问题若极限lim(5an+4bn)=7,极限lim(7an-2bn)=5,则极限lim(6an+bn)=?
若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b
若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
数列极限的题目已知lim(n趋向无穷大)(5n-根号(an^2-bn+c))=2,求a,b的值
等差数列an,bn的前n项和分别为Sn,若Sn/Tn=2n/(3n+1),求lim an/bn
lim(n->无穷)[(3n^2+cn+1)/(an^2+bn)-4n]=5
lim[{根号(n^2+an)}-(bn+1)]=b,求a
数列极限(已知lim[(2n-1)an]=2,求lim n*an)