lim[sinπ/(√n＾2＋1)+sinπ/(√n^2+2)+．．．＋sinπ/√n^2+n),n—>无穷

lim[sinπ/(√n＾2＋1)+sinπ/(√n^2+2)+．．．＋sinπ/√n^2+n),n—>无穷 1、lim(x->无穷大) e^x arctanx2、lim(x->0)sinx√1+sin(1/x)3、lim(x->无穷大)【（√x^2+x+1）-【(√x^2-x+1)】4、lim(x->无穷大)((x+{x+(x)^0.5]^0.5}^0.5)/(2x+1)^0.55、lim(x->0)(sin3x+x^2sin1/x)/((1+cosx)x)6、lim(n->无穷大)(2^n)(si 高数极限lim(n×sin(2π√(n∧2+1))) n→+∞ 计算极限lim（n→∞）{1+ sin[π√(2+4*n^2)]}^n lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? 紧急：求 lim n*sin(π(n^2+2)^0.5）*(-1)^n,n趋向无穷大； 求极限lim(n→∞)sin√(n^2+1)π.可以直接lim(n→∞)sin√(n^2+1)π=sinlim(n→∞)√(n^2+1)π=sinnπ=0吗? lim(n→∞)(sin(n+√(n^2+n)))^2lim(n→∞)(1/n!(1!+2!+…+n!)) 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 求极限 lim Sin[pi*√(n^2+1)] n→∞ 求极限 lim sin pi(n^2+1)^(1/2) 求极限 lim sin pi*(n^2+1)^(1/2) 证明lim （sin√n-sin√n+1）＝0 如何证明lim(sin((n+1)^(1/2))-sin(n^(1/2)))=0 求下列极限 lim（n→∞）∑(上n 下i=1) sin π/(√(n^2+i)) 求极限：lim((2n∧2-3n+1)/n+1)×sin n趋于无穷 高数求极限n趋于无穷大时,lim (1/n - sin(1/n))/ (1/n^2),lim (1/n - sin(1/n))/ (1/n^3)这一式子呢求极限n趋于无穷大时,lim (1/n - sin(1/n))/ (1/n^2),lim (1/n - sin(1/n))/ (1/n^3)这一式子呢? 用定义证明极限lim(sin√(nπ)/(n-√n))