→ craig100 :感謝!! 08/28 17:47
※ 引述《craig100 (不要問,很‧恐‧怖)》之銘言:
: Given that lim g(x)=L ,where L>0
: x->c
: Prove that there exists an open interval (a.b) containing c,
: such that g(x)>0 for all x (- (a,c)∪(c,b)
: 我的做法:
: <proof>:
: for all ε>0 , exist δ>0 such that 0<|x-c|<δ
: => |g(x)-L|<ε
: -δ<x-c<δ
: c-δ< x <c+δ ,and x≠c taking a=c-δ , b=c+δ
: ( c=(b+a)/2,δ=(b-a)/2 )
: -ε<g(x)-L<ε
: L-ε<g(x)<L+ε
: We want to prove L-ε>0 that's done.
: Consider L=g(c)=g((b+a)/2) , δ=(b-a)/2
: L-ε=g((b+a)/2) - (b-a)/2 然後就卡關了.... 不知要怎麼證L-ε>0
: 這樣還有辦法證下去嗎
: 還是有其他證法... 感謝板上高手了!!!
for ε= L/2 > 0 , there exists δ > 0 , s.t. when 0 < │x-c│ < δ
we have │g(x) - L│< L/2
=> -L/2 + L < g(x) < L/2 + L
so g(x) > L/2 > 0 , for all 0 <│x-c│< δ
Hence a = c-δ , b= c+δ are required.
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