n趋向于无穷,[√(n^2＋1)+n]^2/(n^6+1)^3/2.极限等于4是怎么算出来的啊?

n趋向于无穷 求极限lim n趋向于无穷（1/n）*√(n+1)(n+2)⋯(n+n) lim(n/(n^2+1)+...+n/(n^2+n))x趋向于无穷 求解答过程~ lim(2^n+3^n)^1 (n趋向无穷) 求极限(1/n^2)/(e^(1/n)-1-1/n) n趋向于无穷 lim x趋向于无穷 (2^n+1+3^n+1)/(2^n+3^n)= 求lim趋向于无穷3^n+(-2)^n/3^(n+1)+(-2)^(n+1) 当n趋向于无穷时,求n^(1/2)*[n^(1/n)-1]的极限 求极限:Lim(1+1/n-1/n^2)^n n趋向于正无穷 求极限 n趋向于无穷 lim（(根号下n^2+1)/(n+1)）^n 求极限lim(n趋向于无穷）(n+1)(根号下(n^2+1)-n) (1+2^n+3^n)^(1/n) 其中n趋向于正无穷 n趋向于无穷,[√(n^2＋1)+n]^2/(n^6+1)^3/2.极限等于4是怎么算出来的啊? 求极限n趋向于无穷 [(√n+2)-(√n+1)]√n Ps：是根号下的（n+2) 根号下的（n+1） 求极限 lim（ √N^2+N ）－N X趋向于无穷 求极限 n趋向于无穷 lim[1-1/(2n+1)]^n 怎样算 lim(100sin(n!))/(2n-1)n趋向于+无穷.大一高数题. (n+2)^3/(n+1)^4的极限(当n趋向于无穷)