求(3n-sin(n^2))/(2n+cos(n^2))的极限,当n趋于无限大时
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/01 16:19:20
x){I8O83O#/HSS_(O;9{>ټƗ3'<;9Ŷ'M_ xdl)5BΆ* 7h4TxԹ@Y6DP؈\<]G41LCc0Jt;H1
N8=
-yv #
求(3n-sin(n^2))/(2n+cos(n^2))的极限,当n趋于无限大时
求(3n-sin(n^2))/(2n+cos(n^2))的极限,当n趋于无限大时
求(3n-sin(n^2))/(2n+cos(n^2))的极限,当n趋于无限大时
(3n-1)/(2n+1) ≤(3n-sin(n^2))/(2n+cos(n^2))≤ (3n +1)/(2n-1)
lim(n->∞)(3n-1)/(2n+1) ≤lim(n->∞)(3n-sin(n^2))/(2n+cos(n^2))≤ lim(n->∞)(3n +1)/(2n-1)
3/2≤lim(n->∞)(3n-sin(n^2))/(2n+cos(n^2))≤3/2
=> lim(n->∞)(3n-sin(n^2))/(2n+cos(n^2)) =3/2
求极限(sin(2/n)+cos(3/n))^(-n)
求极限(sin^2)n/n+1
求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷
求极限:lim((2n∧2-3n+1)/n+1)×sin n趋于无穷
令n趋近于无穷大,且n存在,求sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=?.
求(3n-sin(n^2))/(2n+cos(n^2))的极限,当n趋于无限大时
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?
如果极限limsin(n)/n=0,则lim)n-3sin(n))/(sin(n)-2n)=______
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
证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1))
组合计算题组合计算题 nC(n-3,n)+P(4,n)=4C(3,n+1)求nC(n+6,3n)+C(3n,n+7) C(17-2n)+C(3n 13+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Lim (2sin^n x+3cos^n x)∕(sin^n x+cos^n x) ,0≤x≤π/2n趋于无穷大
紧急:求 lim n*sin(π(n^2+2)^0.5)*(-1)^n,n趋向无穷大;
n*1+n*2+n*3+n*4.求公式
求最小值 ((n+2)(n+4))/(n(n+3))
(n-3)(n-2)(n-4)等于120 求n
求Sn=C(n,1)+2C(n,2)+...+nC(n,n)C(n,1)+2C(n,2)+...+nC(n,n) n是下标