lim(n^2-n+4)/(2n^2+n-4)=1/2怎么证明用定义可以证明吗?n趋向于无穷

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[(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) 极限练习题lim(n->∞)[(n^2+3n-8)/(4n^2+2n+3)] 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 (n!+(n-1)!+(n-2)!+(N-3)!+⋯..+2!+1)/n!其中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 lim(1-1/n)^(n^2)=? lim(1-1/n^2)^n=?