A nondecreasing sequence which is bounded above converge to the least upper bound of its range pf:Let{an}n=1~infinite be the sequence as above so a1<=a2<=a3<=.... let S={an|1<=n<=infinite} and M=the least upper bound Then we are going to show lim an = M that for epsilon>0 exist k belong to N such that for all n>=k |an-M| <epsilon first since M is an upper bound for all n we have an<=M so for epsilon>0 we have an<M+ epsilon for all n belong to N On the other hand for epsilon >0 by them11.1.2 exist k belong to N such that M-epsilon<ak moreover sine{an} is a nondecreasing sequence M-e<ak<=an n>=k Hence for all n>=k M-e<an so for this epsilon >0 and thos h belong to N we have M-e<an<M+e n>=k Hence we have |an-M|<epsilon ----------------------------------------- Prouduct rule lim an = L , lim bn = M , lim an*bn=L*M pf: for all epsilon>0 exist k1 belong to N n>=k1 such that |an-L|<e/2(|M|-1) for all epsilon>0 exist k2 belont to N n>=k2 such that |an-L|<1 an<|L|+1 for all epsilon>0 exist k3 belong to N n>=k3 such that |bn-M|<e/2(|L|-1) take k=Max{k1,k2,k3} |anbn-LM|=|anbn-anM+anM-LM|<=|an||bn-M|+|M||an-L| <|L+1||bn-M|+|M+1||an-L|=e/2+e/2=e --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.114.218.19 ※ 編輯: winston628 來自: 140.114.218.19 (04/21 01:50) ※ 編輯: winston628 來自: 140.114.218.19 (04/21 01:56)