Question

求极限，各位数学大佬帮帮忙，第二题的an范围怎么求出来

Answer 1

(1)
x->0
(1+x)^(1/2) ~ 1+ (1/2)x -(1/8)x^2
(1-x)^(1/2) ~ 1- (1/2)x -(1/8)x^2
(1+x)^(1/2) +(1-x)^(1/2) -2 ~ -(1/4)x^2
lim(x->0) [(1+x)^(1/2) +(1-x)^(1/2) -2]/x^2
=lim(x->0) -(1/4)x^2/x^2
=-1/4
(2)
a1>1
a(n+1) = 2- 1/an
{an} 是 递减数列
a(n+1) = 2- 1/an
n>1
2- 1/an < an
(an)^2 -2an + 1 >0
an< 1+2√2
c = max { a1, 1+2√2 }
|an| < c

追问

第一题为啥有那个等价呀？老师没讲，第二题如何证明它是单减数列呢？我不是很懂

追答

那是来自泰勒展式

f(x) = f(0)+ [f'(0)/1!]x + [f''(0)/2!]x^2 +....

(1+x)^(1/2) ~ 1+ (1/2)x -(1/8)x^2

(1-x)^(1/2) ~ 1-  (1/2)x -(1/8)x^2 

不懂得话

lim(x->0) [(1+x)^(1/2) +(1-x)^(1/2) -2]/x^2

分子，分母同时乘以  [(1+x)^(1/2) +(1-x)^(1/2) +2] 

=lim(x->0) { (1+x)^(1/2) +(1-x)^(1/2)]^2 -4 }/{ x^2 . [(1+x)^(1/2) +(1-x)^(1/2) +2] }

=(1/4) lim(x->0) { [(1+x)^(1/2) +(1-x)^(1/2)]^2 -4 }/ x^2

=(1/4) lim(x->0) [ (1+x) +2(1-x^2)^(1/2) +(1-x) -4  ]/ x^2

x->0  : (1-x^2)^(1/2) ~ 1- (1/2)x^2

=(1/4) lim(x->0) [ -2 +2(1-x^2)^(1/2)  ]/ x^2

=(1/4) lim(x->0) { -2 +2[1- (1/2)x^2]   }/ x^2                       

=(1/4) lim(x->0)  -x^2/ x^2                       

=-1/4

追问

那数列如何证明它单减呢？

追答

a(n+1) = 2- 1/an
a(n+1) - an
=2- 1/an - an
=-(√an -1/√an)^2
<0

追问

我懂了，谢谢你！