lim 2^n *sin(x/2^n)

Lim(n→∞) 2的n次方sin x/2的n次等于多少? 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/ 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 lim n趋于无穷2的n次方sin(x/2的n次方)的极限怎么求 （n→∞）时lim（2∧n）*(sin(x/2∧n))的极限 紧急：求 lim n*sin(π(n^2+2)^0.5）*(-1)^n,n趋向无穷大； 求极限：lim((2n∧2-3n+1)/n+1)×sin n趋于无穷 求下列极限.lim(n趋向于无穷大)(2x次方)*(sin*1/2x次方) (1)lim(x趋向0)[(1+x)^0.5+(1-x)^0.5-2]/x^2 (2)lim(n趋向无穷)sin[[(n 求极限 lim x-无穷 sin(n+1)/(n+a) 求极限 lim sin pi*(n^2+1)^(1/2) 求极限lim(n->∞)2^n*sin(π/2^n)