Menu Close

Question-158341




Question Number 158341 by aliibrahim1 last updated on 02/Nov/21
Answered by puissant last updated on 03/Nov/21
That is ε>0, let′s seek α>0 / ∀x∈R , 0<∣x−2∣<α ⇒ ∣x^2 −4∣<ε  for x∈R , ∣x^2 −4∣ = ∣x−2∣.∣x+2∣ we have ∣x−2∣<2 → 0<x+2<6  ⇒ ∣x^2 −4∣< 6∣x−2∣ ; ∣x^2 −4∣<ε → 6∣x−2∣<ε → ∣x−2∣<(ε/6)                                                take α=min(2 ; (ε/6))....■                                 .................Le puissant...............
$${That}\:{is}\:\varepsilon>\mathrm{0},\:{let}'{s}\:{seek}\:\alpha>\mathrm{0}\:/\:\forall{x}\in\mathbb{R}\:,\:\mathrm{0}<\mid{x}−\mathrm{2}\mid<\alpha\:\Rightarrow\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid<\varepsilon \\ $$$${for}\:{x}\in\mathbb{R}\:,\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid\:=\:\mid{x}−\mathrm{2}\mid.\mid{x}+\mathrm{2}\mid\:{we}\:{have}\:\mid{x}−\mathrm{2}\mid<\mathrm{2}\:\rightarrow\:\mathrm{0}<{x}+\mathrm{2}<\mathrm{6} \\ $$$$\Rightarrow\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid<\:\mathrm{6}\mid{x}−\mathrm{2}\mid\:;\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid<\varepsilon\:\rightarrow\:\mathrm{6}\mid{x}−\mathrm{2}\mid<\varepsilon\:\rightarrow\:\mid{x}−\mathrm{2}\mid<\frac{\varepsilon}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{take}\:\alpha={min}\left(\mathrm{2}\:;\:\frac{\varepsilon}{\mathrm{6}}\right)….\blacksquare \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……………..\mathscr{L}{e}\:{puissant}…………… \\ $$
Commented by aliibrahim1 last updated on 04/Nov/21
thx sir
$${thx}\:{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *