Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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)

$$\mathrm{Suppose}\:\mathrm{0}<\epsilon<\mathrm{2}\:\mathrm{be}\:\mathrm{arbitrary}\:\mathrm{small}\:\mathrm{numer} \\ $$$$\mathrm{so}\:\mathrm{that}\mid\sqrt{\mathrm{2x}}−\mathrm{2}\mid<\epsilon\Leftrightarrow−\epsilon+\mathrm{2}<\sqrt{\mathrm{2x}}<\epsilon+\mathrm{2} \\ $$$$\Leftrightarrow\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon+\mathrm{4}<\mathrm{2x}<\epsilon^{\mathrm{2}} +\mathrm{4}\epsilon+\mathrm{4} \\ $$$$\Leftrightarrow\frac{\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon+\mathrm{4}}{\mathrm{2}}<\mathrm{x}<\frac{\epsilon^{\mathrm{2}} +\mathrm{4}\epsilon+\mathrm{4}}{\mathrm{2}} \\ $$$$\Leftrightarrow\frac{\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon}{\mathrm{2}}<\mathrm{x}−\mathrm{2}<\frac{\epsilon^{\mathrm{2}} +\mathrm{4}\epsilon}{\mathrm{2}}.\mathrm{Then}\:\mathrm{choosing} \\ $$$$\delta=\frac{\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} }{\mathrm{2}\:}\:\mathrm{we}\:\mathrm{need}\:\mathrm{prove}\:\mathrm{that}\:\forall\mathrm{x}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mid\mathrm{x}−\mathrm{2}\mid<\delta=\frac{\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{then}\mid\sqrt{\mathrm{2x}}−\mathrm{2}\mid<\epsilon.\mathrm{Ineed}, \\ $$$$\mid\mathrm{x}−\mathrm{2}\mid<\delta=\frac{\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} }{\mathrm{2}}\:\Leftrightarrow−\delta+\mathrm{2}<\mathrm{x}<\mathrm{2}+\delta \\ $$$$\Leftrightarrow−\mathrm{2}\delta+\mathrm{4}<\mathrm{2x}<\mathrm{4}+\mathrm{2}\delta \\ $$$$\Rightarrow\epsilon^{\mathrm{2}} −\mathrm{4}\epsilon+\mathrm{4}<\mathrm{2x}<\mathrm{4}\epsilon−\epsilon^{\mathrm{2}} +\mathrm{4}<\mathrm{4}+\mathrm{4}\epsilon+\epsilon^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}−\epsilon<\sqrt{\mathrm{2x}}<\epsilon+\mathrm{2}\Rightarrow\mid\sqrt{\mathrm{2x}−\mathrm{2}}\mid<\epsilon \\ $$$$\mathrm{That}\:\mathrm{shows}\:\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\sqrt{\mathrm{2x}}=\mathrm{2}\left(\boldsymbol{\mathrm{Q}}.\boldsymbol{\mathrm{E}}.\boldsymbol{\mathrm{D}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com