n—>∞时,[sin（1/n^2）]^2等价的无穷小量是ln(1+1/n^4)请问为什么

lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? 当n趋于无穷时,求[sin(π/n)/(n+1)+sin(2π/n)/(n+1/2)+.sinπ/(n+1/n)]的极限 计算极限lim（n→∞）{1+ sin[π√(2+4*n^2)]}^n 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 [1sin(1/n)]/[n²+n+1]+[2sin(2/n)]/[n²+n+2]+……+[nsin(n/n)]/[n²+n+n]求n趋于无穷大时极限? 求极限（sin^2)n/n+1 (sin(ln2/2)+sin(ln3/3)+...+sin(lnn/n))^（1/n）的极限 紧急：求 lim n*sin(π(n^2+2)^0.5）*(-1)^n,n趋向无穷大； n—>∞时,[sin（1/n^2）]^2等价的无穷小量是ln(1+1/n^4)请问为什么 证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1)) lim（1/n）sin n （n→∞） lim[sinπ/(√n＾2＋1)+sinπ/(√n^2+2)+．．．＋sinπ/√n^2+n),n—>无穷 化简sin×[a+(2n+1)π]+2sin×[a-(2n+1)π]/sin(a-2nπ）coS(2nπ-a) (n属于Z) 高数简单求极限lim[(3√n^2)*sin ]/(n+1) n--∞n的3/2次方乘以sin（ n的阶乘） 除以 n+1 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? 为什么sin[pi /(n+√(1+n^2))]当n足够大时 pi /(n+√(1+n^2))∈（0,pi/2）? sin(n*π/2)*sin(n*π/3)*sin(n*π/4)*...*sin(n*π/n-1) 求化简成一个关于n的表达式, lim(n→∞)(sin(n+√(n^2+n)))^2lim(n→∞)(1/n!(1!+2!+…+n!))