Re: [分析] 連續函數一題

[cometic ( )]

※ 引述《handsomecat3 (確定性的失落)》之銘言：
: 設 f(x) is continuous on 開區間 (a,無限大) ，且
: lim (f(x+1)-f(x)) = L
: x->無限大
: 則 lim       f(x)/x  = L                 (Cauchy)
:   x->無限大

Given ε>0, there exist positive integer N (>a) such that

if x≧N then |f(x+1)-f(x)-L|<ε/4



Since f(x) continuous on [N,N+1], so there exist B>0 such that

|f(x)|≦B for all x belong to [N,N+1]



Since limit B/x =0 ,limit L(([x]-N)/x-1)=0 and limit ([x]-N)/x=1
      x->∞         x->∞                      x->∞

([x] is the largest integer not greater than x)

, so there exist M (>N+1) such that

if x≧M then |B/x|< ε/4 and |L(([x]-N)/x-1)|< ε/4 and 0< ([x]-N)/x <2



Let g(x)= x-[x]+N if x≧N

Clearly, we have

(1)
g(x)≧N if x≧N

(2)
x=g(x)+[x]-N if x≧N



If x>M: (==> x>N+1, g(x)≧N, [x]≧N+1)

then

L-ε/4<    f(g(x)+1) -   f( g(x) )    <L+ε/4
L-ε/4<    f(g(x)+2) -   f(g(x)+1)    <L+ε/4
                     .
                     .
                     .
L-ε/4< f(g(x)+[x]-N)-f(g(x)+[x]-N-1) <L+ε/4

==>
(L-ε/4)([x]-N)    < f(g(x)+[x]-N)-f(g(x)) <(L+ε/4)([x]-N)

(L-ε/4)([x]-N)/x  <   (f(x)-f(g(x)))/x    <(L+ε/4)([x]-N)/x

(L-ε/4)([x]-N)/x-L<  (f(x)-f(g(x)))/x-L   <(L+ε/4)([x]-N)/x-L


Since x>M, so

(L-ε/4)([x]-N)/x-L=L(([x]-N)/x-1)-ε/4([x]-N)/x
                   >-ε/4-(ε/4)*2=-(3/4)ε

(L+ε/4)([x]-N)/x-L=L(([x]-N)/x-1)+ε/4([x]-N)/x
                   <ε/4+(ε/4)*2=(3/4)ε

==>

         - (3/4)ε < (f(x)-f(g(x)))/x-L < (3/4)ε

-(3/4)ε+f(g(x))/x <      f(x)/x-L      < (3/4)ε+f(g(x))/x

-(3/4)ε-B/x       <      f(x)/x-L      < (3/4)ε+B/x

|f(x)/x-L|<(3/4)ε+B/x<(3/4)ε+ε/4=ε


So

limit f(x)/x =L
x->∞

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※ 編輯: cometic         來自: 140.114.203.189      (08/05 15:13)
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