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討論串[分析] uniformly continuous
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Definition od Lipschitz condition:. http://planetmath.org/encyclopedia/Lipschitzcondition.html. http://mathworld.wolfram.com/LipschitzCondition.html.
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不對. 要對任意ε. given ε>0. since g'→0 as x→∞. choose M such that ∣g'(x)∣<ε if x>M. for x>y >M choose δ= 1. if ︱x-y︱<δ then ∣g(x)-g(y)∣=∣x-y∣∣g'(z)∣for some
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關於 g(x)=e^(-x) , x屬於[0,∞) 驗證g為uni. continuity. 不知這樣寫可否. let ε=1 ,δ>0 必存在r屬於正整數 s.t. rδ^2 > 1-δ. let x=ln(rδ^2) y=ln(rδ). x-y = ln(rδ^2)-ln(rδ) = lnδ <
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Since lim f(x) = L. we have for all ε>0, there exists M > 0 s.t.. for all x ≧ M, |f(x) - L| < ε. 1.for xε[0,M], f(x) is uniformly continuous.. (contin
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(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,
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