2^n.sinπ/2^n (n趋近无穷),求极限,用两个重要极限公式求
2^n.sinπ/2^n (n趋近无穷),求极限,用两个重要极限公式求
求极限 X趋近于无穷 2^n(sin(x/2^n))
(5^n+(-2)^n)/(5^(n+1)+(-2)^(n+1))当n趋近无穷,求极限.
求极限 lim n趋近无穷 (n+1)^2/2n^2-3
求极限lim((n+1)/(n2+1)+(n+2)/(n2+2)+...+(n+n)/(n2+n)),n趋近无穷
极限 limn趋近于正无穷(2^n-3^n)/4^n如何求呀?
求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷
求极限lim(n趋近无穷){1/(n^2+1)+2/(n^2+1)+...+2n/(n^2+1)}
2^n/n,n趋近无穷大的极限怎么求?
求极限:lim((2n∧2-3n+1)/n+1)×sin n趋于无穷
n趋于正无穷求极限n^2*ln[n*sin(1/n)]
求极限 n*sin(x/n) n趋向无穷