若极限lim((3n^2+4)/(n+2)-an+b)=6,则a=_,b=_

若极限lim((3n^2+4)/(n+2)-an+b)=6,则a=_,b=_ lim (n→∞) (n^2/(an+b)-n^3/(2n^2-1))=1/4 求a,b 若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b 若lim[(an^2+bn+c)/(2n-3)]=-2,则a+b= 求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 求a b的值 几道大一 求极限：1.lim(√n -9)/(n+3)=2.lim(1+a+a^2+a^3+.+a^n)/(1+b+b^ 数列极限基本题已知数列{an}的极限为0,且有lim[(3n-2)an]=6,则lim[n(an)]=? lim (n→∞) [(an^2+bn+c)/(2n+5)]=3,求a,b 已知lim[(an^2+5n-2)/(3n+1)-n]=b,求a,b的值 (n²+1)/(n+1)-an-b的极限为0,则a+2b= lim（n2+2n+2）/（n+1）-an）=b,求a,b 已知：lim (n→∞) [(n^2+n)/(n+1)-an-b]=1 ,求a,b的值