lim (1+2/n)^n+4 n-->无穷大 求极限

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/n^2+4/n^2+7/n^2+…+3n-1/n^2) 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 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->∞ lim[(n+3)/(n+1))]^(n-2) 【n无穷大】 lim(2^n+3^n)^1 (n趋向无穷) 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(1-1/n)^(n^2)=? lim(1-1/n^2)^n=? lim(n→∞) 1/(n+1)-2/(n+1)+3/(n+1)-4/(n+1)+...+[(2n-1)/(n+1)]-[(2n)/(n-1)]求极限 求极限,lim(1+n)(1+n^2)(1+n^4)-----(1+n^2n)=?（n趋于无穷） lim{[n*(n+1)*……*(2n-1)]^1/n}/n n->无穷答案是4/e