lim(1+2+4+...+2^n-1)/(1+3+9+...+3^n-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? lim(1-1/n)^(n^2)=? lim(1-1/n^2)^n=? lim[(n+3)/(n+1))]^(n-2) 【n无穷大】 lim(2^n+3^n)^1 (n趋向无穷) lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2) 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 In(n+1) / In(n+2) 是多少lim [In(n+1) / In(n+2)] 是多少 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) 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 求1.lim(3n-(3n^2+2n)/(n-1)) 2.lim(8+1/(n+1)) 3.lim根号n(根号(n+1)-根号(n-3))