pure (URGENT)
2010-10-14 4:05 am
回答 (1)
2010-10-17 2:10 am
✔ 最佳答案
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2010-10-16 18:10:39 補充：
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2010-10-17 14:51:02 補充：
Q3. Correction. (a) If x_k>0 then x_(k+1)>0 =>x_n >0
(b) lim y_(n-1) < k y_n
Suppose the limit is L
L <= kL
L(1-k) <=0
Since L >=0, hence L=0 => lim x_n = sqrt(3)
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101013000051KK01174