lim->无穷(2n^2/(n+2)-an)=b,求常数a,b的值

若lim n->无穷 [ 2n^2/(n+2) - an ]=b,求常数a,b的值 求过程,感谢.. a,b为常数.lim（n->无穷）an^2+bn+2/2n-1=3 求a,b 已知lim n→无穷 （an^2+bn+5）/(3n-2)=2,求a,b的值 已知：lim (n→∞) [(n^2+n)/(n+1)-an-b]=1 ,求a,b的值 已知lim[(3n^2+cn+1)/(an^2+bn)-4n]=5,求常数a、b、c的值 已知lim[(an^2+5n-2)/(3n+1)-n]=b,求a,b的值 lim[(n^2+1)/(n+1)-an-b]=o求a和b的值 lim[1/(a-1)^n]=0 求a的范围 lim n->无穷 (1+a+a^2+...+a^n)/(1+b+b^2+...+b^n) lim（n2+2n+2）/（n+1）-an）=b,求a,b 设f(x)=lim n→正无穷[x^(2n-1)+ax^2+bx]/(x^2n+1)是连续函数,求a,b的值 lim (n→∞) (n^2/(an+b)-n^3/(2n^2-1))=1/4 求a,b 若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值