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一題極限證明題(有點難度)
f is continuous on (0,∞) 0<a<b<∞, and for all h in (a,b)
f(nh)-->0 as n-->∞
then f(x)-->0 as x-->∞
2 個解答
- 1 十年前最愛解答
f(nh)-->0 as n-->∞
由這句話知道
f(x)大概是以下這種形式:
f(x)=a0/(x+b0) + a1/[(x+b1)^2+c1] + ......
an/[(x+bn)^n+cn]
如此一來便不用管h的大小了
當然f(x)-->0 as x-->∞ 也會成立
- 約翰教練Lv 41 十年前
Given an h such that 0<a<h<b<infinite
By the condition , given epsilon>0 , there exists N such that | f(nh) |<epsilon , whenever n > N
And by the same epsilon and N , we have that | f(nh) | <epsilon , whenever (nh) > (Nh)
Choose Nh = N' , then for the epsilon be given , there exists N' such that | f(x) | < epsilon , whenever x > N'
Since nh---> infinite as n---> infinite , for all h in the segment which is given , we can choose {x}=nb such that x is independent on h and | f(x) | < | f(nh) | < epsilon ,for x is large enough
