Re: [分析] uniformly continuous
(lim means lim )
x->∞
Given ε>0.
Since lim f(x) = L, there exist M > 0 such that
for x ≧ M, |f(x) - L| < ε/2.
Consider [0,M+1] is a compact set, there exists δ1>0 such that
|f(x)-f(y)| < ε whenever |x-y| < δ1.
Let δ=min{δ1,1/2}
If |x-y| < δ, WLOG we may assume, x > y; then there are two cases:
(i) y ≧M
(ii) y < M.
For the first case, x>y≧M, so |f(x)-L|<ε/2 and |f(y)-L|<ε/2,
hence |f(x)-f(y)| < ε
For the second case, sine x-y<1/2, so x < 1/2 + y ≦ 1 + M.
So, x,y are in [0,M+1]; consdier |x-y|< δ≦δ1,
|f(x)-f(y)|< ε.
Hence f is uniformly continuous on [0,oo)
※ 引述《VFresh (車干)》之銘言：
: ※ 引述《wyob (Go Dolphins)》之銘言：
: : If f:[0,∞)→R be a continuous function. Assume that limf(x)=L
: : x→∞
: : where L is a finite number.Show that f is unifomly continuous on [0,∞)
: : 感謝解惑
: 在此給的想法 建議原PO往此方向做做看
: lim f = L , 代表 x 夠大之後 f(x) 會與 L 很接近
: 換句話說 |f(x)-L|就會很小
: 因此對於所有 x,y 夠大之後 |f(x)-L|, |f(y)-L|就會很小
: 因此 |f(x)-f(y)|也會很小
: 那x 不夠大的時候呢? 考慮 前段是個 closed bounded interval
: (因此是個compact set) 此函數連續 因此在此均勻連續.
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◆ From: 114.25.53.139
※ 編輯: VFresh 來自: 114.25.53.139 (01/08 19:30)
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