lim 2^n *sin(x/2^n)n→∞求极限

Lim(n→∞) 2的n次方sin x/2的n次等于多少? lim 2^n *sin(x/2^n)n→∞求极限 lim(n→∞){2^n[sin(x/2^n)]}求详细过程,谢谢! 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? 求Lim (2sin^n x+3cos^n x)∕(sin^n x+cos^n x) ,0≤x≤π/2n趋于无穷大 如果极限limsin(n)/n=0,则lim)n-3sin(n))/(sin(n)-2n)=______ lim x趋向于无穷 2^n sin(x/2^n) 求极限Lim（(n-x)/(n+2))^(n+1) lim(n→∞) (cos x/n)^n^2 求极限：lim(sinx/(2^n*sin(x/2^n)))要详细过程, x→∞ lim(5n/2)sin(2π/n) lim n趋近无穷大n^2*(2-n*sin 2/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((2n∧2-3n+1)/n+1)×sin n趋于无穷 lim n趋向于无穷 2^n sin(x/2^n)X为不等于0的常数 lim n趋于无穷2的n次方sin(x/2的n次方)的极限怎么求 利用无穷小的性质求极限lim(x→+∞)[(n^2+1)/n^3]sin(n!)=