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Question Number 116819 by joki last updated on 07/Oct/20
prove the limit lim_(x−⟩2) (√(2x))=2
$${prove}\:{the}\:{limit} \\ $$$${li}\underset{{x}−\rangle\mathrm{2}} {{m}}\sqrt{\mathrm{2}{x}}=\mathrm{2} \\ $$
Answered by 1549442205PVT last updated on 07/Oct/20
Suppose 0<ε<2 be arbitrary small numer so that∣(√(2x))−2∣<ε⇔−ε+2<(√(2x))<ε+2 ⇔ε^2 −4ε+4<2x<ε^2 +4ε+4 ⇔((ε^2 −4ε+4)/2)<x<((ε^2 +4ε+4)/2) ⇔((ε^2 −4ε)/2)<x−2<((ε^2 +4ε)/2).Then choosing δ=((4ε−ε^2 )/(2 )) we need prove that ∀x so that ∣x−2∣<δ=((4ε−ε^2 )/2) then∣(√(2x))−2∣<ε.Ineed, ∣x−2∣<δ=((4ε−ε^2 )/2) ⇔−δ+2<x<2+δ ⇔−2δ+4<2x<4+2δ ⇒ε^2 −4ε+4<2x<4ε−ε^2 +4<4+4ε+ε^2 ⇒2−ε<(√(2x))<ε+2⇒∣(√(2x−2))∣<ε That shows lim_(x→2) (√(2x))=2(Q.E.D)
$$\mathrm{Suppose}\:\mathrm{0}<\epsilon<\mathrm{2}\:\mathrm{be}\:\mathrm{arbitrary}\:\mathrm{small}\:\mathrm{numer} \\ $$$$\mathrm{so}\:\mathrm{that}\mid\sqrt{\mathrm{2x}}−\mathrm{2}\mid<\epsilon\Leftrightarrow−\epsilon+\mathrm{2}<\sqrt{\mathrm{2x}}<\epsilon+\mathrm{2} \\ $$$$\Leftrightarrow\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon+\mathrm{4}<\mathrm{2x}<\epsilon^{\mathrm{2}} +\mathrm{4}\epsilon+\mathrm{4} \\ $$$$\Leftrightarrow\frac{\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon+\mathrm{4}}{\mathrm{2}}<\mathrm{x}<\frac{\epsilon^{\mathrm{2}} +\mathrm{4}\epsilon+\mathrm{4}}{\mathrm{2}} \\ $$$$\Leftrightarrow\frac{\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon}{\mathrm{2}}<\mathrm{x}−\mathrm{2}<\frac{\epsilon^{\mathrm{2}} +\mathrm{4}\epsilon}{\mathrm{2}}.\mathrm{Then}\:\mathrm{choosing} \\ $$$$\delta=\frac{\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} }{\mathrm{2}\:}\:\mathrm{we}\:\mathrm{need}\:\mathrm{prove}\:\mathrm{that}\:\forall\mathrm{x}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mid\mathrm{x}−\mathrm{2}\mid<\delta=\frac{\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{then}\mid\sqrt{\mathrm{2x}}−\mathrm{2}\mid<\epsilon.\mathrm{Ineed}, \\ $$$$\mid\mathrm{x}−\mathrm{2}\mid<\delta=\frac{\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} }{\mathrm{2}}\:\Leftrightarrow−\delta+\mathrm{2}<\mathrm{x}<\mathrm{2}+\delta \\ $$$$\Leftrightarrow−\mathrm{2}\delta+\mathrm{4}<\mathrm{2x}<\mathrm{4}+\mathrm{2}\delta \\ $$$$\Rightarrow\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon+\mathrm{4}<\mathrm{2x}<\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} +\mathrm{4}<\mathrm{4}+\mathrm{4}\epsilon+\epsilon^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}−\epsilon<\sqrt{\mathrm{2x}}<\epsilon+\mathrm{2}\Rightarrow\mid\sqrt{\mathrm{2x}−\mathrm{2}}\mid<\epsilon \\ $$$$\mathrm{That}\:\mathrm{shows}\:\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\sqrt{\mathrm{2x}}=\mathrm{2}\left(\boldsymbol{\mathrm{Q}}.\boldsymbol{\mathrm{E}}.\boldsymbol{\mathrm{D}}\right) \\ $$

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