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Question Number 116939 by joki last updated on 08/Oct/20
prof  if  the limitf(x)=L and   limit f(x)=M,then L=M
profifthelimitf(x)=Landlimitf(x)=M,thenL=M
Answered by MAB last updated on 08/Oct/20
lim_(x→x_0 ) f(x)=L=M  ⇒[(∀ε>0)(∃α>0)(∀x∈D_f ):∣x−x_0 ∣<α⇒∣f(x)−L ∣<ε]  and [(∀ε′>0)(∃α′>0)(∀x∈D_f ):∣x−x_0 ∣<α′⇒∣f(x)−M ∣<ε′]  we have ∣M−L∣≤∣f(x)−L∣+f(x)−M∣ triangle inequality  hence (∀ε>0)(∃δ=min(α,α′)>0): ∣x−x_0 ∣<δ⇒∣M−L∣<∍  finally M=L
limxx0f(x)=L=M[(ε>0)(α>0)(xDf):∣xx0∣<α⇒∣f(x)L∣<ε]and[(ε>0)(α>0)(xDf):∣xx0∣<α⇒∣f(x)M∣<ε]wehaveML∣⩽∣f(x)L+f(x)Mtriangleinequalityhence(ϵ>0)(δ=min(α,α)>0):xx0∣<δ⇒∣ML∣<∍finallyM=L
Answered by MAB last updated on 08/Oct/20
lim_(x→x_0 ) f(x)=L=M  ⇒[(∀ε>0)(∃α>0)(∀x∈D_f ):∣x−x_0 ∣<α⇒∣f(x)−L ∣<ε]  and [(∀ε′>0)(∃α′>0)(∀x∈D_f ):∣x−x_0 ∣<α′⇒∣f(x)−M ∣<ε′]  we have ∣M−L∣≤∣f(x)−L∣+f(x)−M∣ triangle inequality  hence (∀ε>0)(∃δ=min(α,α′)>0): ∣x−x_0 ∣<δ⇒∣M−L∣<∍  finally M=L
limxx0f(x)=L=M[(ε>0)(α>0)(xDf):∣xx0∣<α⇒∣f(x)L∣<ε]and[(ε>0)(α>0)(xDf):∣xx0∣<α⇒∣f(x)M∣<ε]wehaveML∣⩽∣f(x)L+f(x)Mtriangleinequalityhence(ϵ>0)(δ=min(α,α)>0):xx0∣<δ⇒∣ML∣<∍finallyM=L

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