lim(2+4+6+...+2^n)/(1+3+5+...+2^n-1)lim(1+1/2+1/4+...+1/2^n-1)/(1+1/3+1/9+...+1/3^n-1)怎么解n-无穷大

lim(n+3)(4-n)/(n-1)(3-2n) lim(n^3+n)/(n^4-3n^2+1) lim[(4+7+...+3n+1)/(n^2-n)]= lim (1+2/n)^n+4 n-->无穷大 求极限 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? 大一微积分解答：lim(n→+∞)（2^n+4^n+6^n+8^n）^1/n=? 求极限,lim(x->0) (1-2sinx)^(3/x)lim(n->+∞) (n!-4^n) / (6+ln(n)+n^2) 计算下列极限：1）lim(n→∞) 1/n3 2）lim(n→∞)4n+1/3n-11）lim(n→∞) 1/n3 2）lim(n→∞)4n+1/3n-13) lim(n→∞) (1/3)n4)lim(n→∞)n3+2n-5/5n3-n 5) lim(n→∞)(1+1/2n)n 6) lim(n→∞)2x3-x2+1/3x2+2x-9 7) lim(x→0 )sin3x/sin7x8) lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2) 极限练习题lim(n->∞)[(n^2+3n-8)/(4n^2+2n+3)] 求极值 lim(n->无限) [n(n+1)(2n+1)]/6n^2如题,.. lim n趋向无穷大3n^3+n^2-3/4n^3+2n+1 求lim(n+1)(n+2)(n+3)/(n^4+n^2+1) 一道极限题,lim[n^2(2n+1)]/(n^3+n+4)n->∞ 1、lim n->无穷 根号[(n^4+n+1)-n^2]*(3n+4) lim(n→∞) (3^n-4^n)/(3^n+2×4^n),请计算, lim(2+4+6+...+2n)/(1+3+5+...+(2n-1)) 几道数列极限的题目(要过程)化简1) lim an平方+bn+c/2n-3=-2 则a+b=?2) lim (1+3/n)的n次方3)lim(1+2/n+2)的n次方4)lim 1+2+4+6+.+2的n次方/2n次方lim下面有个n->无穷大 懂得人知道就好 过程中lim就不用写了!尽快!