设 _(1)=10, _(n+1)=sqrt (6+{a)_(n)} 证明:极限liman存在,并求之.
6.设x_(1)=sqrt(6),x_(n+1)=sqrt(6+x_(n))(n=1,2,...),证明数列x_{n)}收敛,并求出极限值.6.设$x_{1}=
+ X_(10)^2))服从t分布,自由度为() A (2)/(sqrt(6)), 6 B (sqrt(6))/(2), 6 C (sqrt(
.求下列极限:-|||-(6) lim _(xarrow 4)dfrac (sqrt {2x+1)-3}(sqrt {x-2)-sqrt (2)} ;
(5)dfrac (6+2sqrt {3)}(2)
求下列极限:-|||-(6) lim _(xarrow +infty )(x)^dfrac (3{2)}(sqrt (x+2)-2sqrt (x+1)+sqrt
下列复数① sqrt[4](6)(cos (pi)/(8)+i sin (pi)/(8));② sqrt[4](6)(cos (pi)/(8)-i sin (p
求下列函数的极限。-|||-lim _(xarrow +infty )x(3x-sqrt (9{x)^2-6})
计算:(1)(sqrt(5))2;(2)(-sqrt(0.2))2;(3)(sqrt((2)/(7)))2;(4)(5sqrt(5))2;(5)sqrt((-1
3.计算下列极限:-|||-(6) lim _(narrow infty )(sqrt ({n)^2+n}-sqrt ({n)^2-n});