高微(definition of convergence)

2012-09-19 3:32 am

Let {x_n} be a sequence of real numbers, and suppose that x_n → -1 as n →∞.
Show that
(a) ((x_n)^2 - √3)/x_n → -1+√3 as n → ∞.
(b) Show that there exists a constant C<0 and an integer N∈N such that
x_n<C<0 whenever n≥N.
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這裡的(a)雖然一看就知道了，但是我想用definition of convergence去做
x_n → -1 as n →∞:
∀ε>0 ∃N∈N, s.t. if n≥N, then |(x_n)+1|<ε
然後原來的式子
| ((x_n)^2 - √3)/x_n - (-1+√3) |=|(x_n-√3)/x_n|*|(x_n)+1|, 做到這裡就停下來了
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請大大們幫忙，謝謝~!

更新1:

大大你好 There exists N1 such that whenever n≧N1 then |x_n-x|<|x|/2, so |x_n|>|x|/2 |x_n|>|x|/2 是如何推導出來的?

回答 (1)

老怪物

2012-09-19 5:30 am

✔ 最佳答案

(a)
x_n→x≠0, then (x_n^2-a)/x_n→(x^2-a)/x.

Proof:

|(x_n^2-a)/x_n - (x^2-a)/x| = |(x_n-a/x_n)-(x-a/x)|
≦ |x_n-x| + |a|．|1/x_n-1/x|
= |x_n-x| + |a|．|x_n-x|/(|x_n|．|x|)

There exists N1 such that
whenever n≧N1 then |x_n-x|<|x|/2, so |x_n|>|x|/2

For any ε>0 there exists N2 such that
whenever n≧N2 then |x_n-x|<ε.

Therefore, whenever n≧N=max{N1,N2}, then
|(x_n^2-a)/x_n - (x^2-a)/x| < ε + |a|ε/(|x|^2/2) = (1+2|a|/x^2)ε.

將 a=√3, x=-1 代入以上論證, 即得 (x_n^2-√3)/x_n→(1-√3)/(-1).

(b)
x_n→x<0, 則存在 C<0 及 N such that
whenever n≧N then x_n<C.

Proof:
Take C=x/2, 則 C<0.
存在 N such that whenever n≧N then |x_n-x|<C-x = -x/2
故 n≧N implies x_n<x+(C-x) = C.

2012-09-19 10:16:18 補充：
|x_n-x|<|x|/2 ==> ||x_n|-|x||<|x|/2 ==> |x_n|>|x|-|x|/2=|x|/2.

或者, 依 x>=0 與 x<0 兩種情況拆解不等式 |x_n-x|<|x|/2.

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