lim(2+4+6+...+2n)/(1+3+5+...+(2n-1))

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(1-1/n)^(n^2)=? lim(1-1/n^2)^n=? lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2) 求极值 lim(n->无限) [n(n+1)(2n+1)]/6n^2如题,.. 求lim(n+1)(n+2)(n+3)/(n^4+n^2+1) 1、lim n->无穷 根号[(n^4+n+1)-n^2]*(3n+4) lim n趋向无穷大3n^3+n^2-3/4n^3+2n+1 一道极限题,lim[n^2(2n+1)]/(n^3+n+4)n->∞ 求极限 lim(n→∞) tan^n (π/4 + 2/n) lim(n→∞)tan^n(π/4+2/n) =lim(n→∞)[(tan(π/4)+tan(2/n))/(1-tan(π/4)tan(2/n))]^n =lim(n→∞)[(1+tan(2/n))/(1-tan(2/n))]^n =lim(n→∞)(1+tan(2/n))^n/(1-tan(2/n))^n (1) 因为 lim(n→∞)(1+tan(2/n) lim[(n+3)/(n+1))]^(n-2) 【n无穷大】