lim(3n+4n+5n)1/n（那是3的n次幂+4的n次幂+5的n次幂）的1/n次幂,n趋向于无穷大

lim(n+3)(4-n)/(n-1)(3-2n) lim(n^3+n)/(n^4-3n^2+1) lim(1/n+2/n+3/n+4/n+5/n+……+n/n)=lim(1/n)+lim(2/n)+……+lim(n/n)成立吗?（n趋近于无穷大）为什么不成立? lim(3n+4n+5n)1/n（那是3的n次幂+4的n次幂+5的n次幂）的1/n次幂,n趋向于无穷大 lim(3^2n+5^n)/(1+9^n)=? lim[(4+7+...+3n+1)/(n^2-n)]= 用数列极限证明lim(n→∞)(n^-2)/(n^+n+1)=1中证明如下:lim(n→∞)3n+1/5n-4 lim((2n+1)^5(n+10)^3)/((n^2-2n-3)^4)=32 求lim(n->无穷)(3n+5)/(n^2+n+4)^1/2的极限 lim[(n+3)/(n+1))]^(n-2) 【n无穷大】 lim(2^n+3^n)^1 (n趋向无穷) lim n趋向无穷大3n^3+n^2-3/4n^3+2n+1 lim(3^2n+5^n)/(1+9^n)3^2n or 5^n 1、lim n->无穷 根号[(n^4+n+1)-n^2]*(3n+4) 求lim(n+1)(n+2)(n+3)/(n^4+n^2+1) lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2) 一道极限题,lim[n^2(2n+1)]/(n^3+n+4)n->∞ 为什么lim n趋向无穷 (5n^2-4n)/(3n^2+2n-1)等于5/3