求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 求a b的值
求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 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
若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
已知数列{an}{bn}满足a1=1,a2=3,b(n+1)/bn=2,bn=a(n+1)-an,(n∈正整数),求数列
数列极限(已知lim[(2n-1)an]=2,求lim n*an)
已知lim[(3n^2+cn+1)/(an^2+bn)-4n]=5,求常数a、b、c的值
lim[{根号(n^2+an)}-(bn+1)]=b,求a
a,b为常数.lim(n->无穷)an^2+bn+2/2n-1=3 求a,b
1 lim[2n+1-根号(an^2+bn+1)]=2 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
已知lim n→无穷 (an^2+bn+5)/(3n-2)=2,求a,b的值