lim n〔√(n^2+1)-n〕当n→∞时的极限

lim n〔√(n^2+1)-n〕当n→∞时的极限 求当n→∞,Lim(1+2+3+4+……+(n-1)+n)/n 用夹逼定理求lim(n→∞)[√(n^2+n)-n]^(1/n) 用夹逼定理求lim(n→∞)√[(n^2+n)-n]^(1/n) lim x→n (√n+1-√n)*√(n+1/2)lim x n→∞ (√n+1-√n)*√(n+1/2) lim[n/(n*n+1*1)+n/(n*n+2*2)+...+n/(n*n+n*n)],当x趋向无穷大时,怎么求极限, lim (n!+(n-1)!+(n-2)!+(N-3)!+⋯..+2!+1)/n!其中n→∞ lim（n→∞) （(2n!/n!*n)^1/n的极限用定积分求是lim（n→∞) 1/n(2n!/n!)^1/n 不好意思 lim(n→∞)(sin(n+√(n^2+n)))^2lim(n→∞)(1/n!(1!+2!+…+n!)) lim n→∞ n/√(2n+1)(n+1)= 极限lim(n→∞)√(n^2-3n)/2n+1 等于多少 lim √(n^2-3n)/2n+1 n→0 求lim n→∞ (1+2/n)^n+3 lim(n→∞)[1-(2n/n+3)] lim(n→∞)(2n-1/n+3) lim[1/(n+1)+1/(n+2)+1/(n+3)+、、、1/(n+n)]当n趋于无穷大时的极限? lim(n+3)(4-n)/(n-1)(3-2n) lim[(n+3)/(n+1))]^(n-2) 【n无穷大】