(√n sin n )/ (n+1) 求当n趋近正无穷的时候的极限.

求极限（sin^2)n/n+1 (√n sin n )/ (n+1) 求当n趋近正无穷的时候的极限. 求sin(√(n+1))-sin(√n),当n趋向无穷的极限 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 求极限,lim n趋向无穷大,√n*sin(1/√n)=1 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 当n趋于∞,[(n+1)^(n+1)/n^n]sin(1/n)的极限怎么求 极限 lim(x-->无穷)(1/n sin(n)+1/n sin(1/n)+nsin(1/n))求极限lim(x-->无穷)(1/n sin(n)+1/n sin(1/n)+nsin(1/n)) [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：lim((2n∧2-3n+1)/n+1)×sin n趋于无穷 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→∞ 根号n乘以sin n 除以n+1 求∑sin(n^2+1)π/n条件收敛 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 |sin(1/n)| 求sinπ/n*sin2π/n*…*sin(n-1)π/n的值,用复数思想 求极限(sin(2/n)+cos(3/n))^(-n) 求极限 n*sin(x/n) n趋向无穷