Question Number 116939 by joki last updated on 08/Oct/20
$${prof}\:\:{if}\:\:{the}\:{limitf}\left({x}\right)={L}\:{and}\: \\ $$$${limit}\:{f}\left({x}\right)={M},{then}\:{L}={M} \\ $$
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](Q116944.png)
$$\underset{\boldsymbol{\mathrm{x}}\rightarrow{x}_{\mathrm{0}} } {\boldsymbol{\mathrm{lim}}}{f}\left({x}\right)={L}={M} \\ $$$$\Rightarrow\left[\left(\forall\varepsilon>\mathrm{0}\right)\left(\exists\alpha>\mathrm{0}\right)\left(\forall{x}\in{D}_{{f}} \right):\mid{x}−{x}_{\mathrm{0}} \mid<\alpha\Rightarrow\mid{f}\left({x}\right)−{L}\:\mid<\varepsilon\right] \\ $$$${and}\:\left[\left(\forall\varepsilon'>\mathrm{0}\right)\left(\exists\alpha'>\mathrm{0}\right)\left(\forall{x}\in{D}_{{f}} \right):\mid{x}−{x}_{\mathrm{0}} \mid<\alpha'\Rightarrow\mid{f}\left({x}\right)−{M}\:\mid<\varepsilon'\right] \\ $$$${we}\:{have}\:\mid{M}−{L}\mid\leqslant\mid{f}\left({x}\right)−{L}\mid+{f}\left({x}\right)−{M}\mid\:{triangle}\:{inequality} \\ $$$${hence}\:\left(\forall\epsilon>\mathrm{0}\right)\left(\exists\delta={min}\left(\alpha,\alpha'\right)>\mathrm{0}\right):\:\mid{x}−{x}_{\mathrm{0}} \mid<\delta\Rightarrow\mid{M}−{L}\mid<\backepsilon \\ $$$${finally}\:{M}={L} \\ $$
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](Q116945.png)
$$\underset{\boldsymbol{\mathrm{x}}\rightarrow{x}_{\mathrm{0}} } {\boldsymbol{\mathrm{lim}}}{f}\left({x}\right)={L}={M} \\ $$$$\Rightarrow\left[\left(\forall\varepsilon>\mathrm{0}\right)\left(\exists\alpha>\mathrm{0}\right)\left(\forall{x}\in{D}_{{f}} \right):\mid{x}−{x}_{\mathrm{0}} \mid<\alpha\Rightarrow\mid{f}\left({x}\right)−{L}\:\mid<\varepsilon\right] \\ $$$${and}\:\left[\left(\forall\varepsilon'>\mathrm{0}\right)\left(\exists\alpha'>\mathrm{0}\right)\left(\forall{x}\in{D}_{{f}} \right):\mid{x}−{x}_{\mathrm{0}} \mid<\alpha'\Rightarrow\mid{f}\left({x}\right)−{M}\:\mid<\varepsilon'\right] \\ $$$${we}\:{have}\:\mid{M}−{L}\mid\leqslant\mid{f}\left({x}\right)−{L}\mid+{f}\left({x}\right)−{M}\mid\:{triangle}\:{inequality} \\ $$$${hence}\:\left(\forall\epsilon>\mathrm{0}\right)\left(\exists\delta={min}\left(\alpha,\alpha'\right)>\mathrm{0}\right):\:\mid{x}−{x}_{\mathrm{0}} \mid<\delta\Rightarrow\mid{M}−{L}\mid<\backepsilon \\ $$$${finally}\:{M}={L} \\ $$