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^2+1)/(n+1)-an-b]=o求a和b的值 lim[1/(a-1)^n]=0 求a的范围
已知:lim (n→∞) [(n^2+n)/(n+1)-an-b]=1 ,求a,b的值
已知lim[(an^2+5n-2)/(3n+1)-n]=b,求a,b的值
lim(n2+2n+2)/(n+1)-an)=b,求a,b
若lim(2n+(an^2-2n+1)/(bn+2))=1 求a/b的值
lim (n→∞) (n^2/(an+b)-n^3/(2n^2-1))=1/4 求a,b
已知lim(n的平方+1/n+1+an-b)=1,求a,b的值
已知lim((an2+5n-2)/(3n+1) -n)=b 求a b的值
lim(n2+1/n+1-an-b)=0,求a,b
a,b为常数.lim(n->无穷)an^2+bn+2/2n-1=3 求a,b
求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 求a b的值
求极限的问题:lim(n→∞) {[a^(1/n)+b^(1/n)/2} 其中a,b大于0