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Question Number 116819 by joki last updated on 07/Oct/20

$$provethelimit$$$$li\underset{x-⟩2}{m}\sqrt{2x}=2$$

Answered by 1549442205PVT last updated on 07/Oct/20

$$\mathrm{Suppose}0<ϵ<2\mathrm{be}\mathrm{arbitrary}\mathrm{small}\mathrm{numer}$$$$\mathrm{so}\mathrm{that}\mid \sqrt{2\mathrm{x}}-2\mid <ϵ\iff -ϵ+2<\sqrt{2\mathrm{x}}<ϵ+2$$$$\iff {ϵ}^2-4ϵ+4<2\mathrm{x}<{ϵ}^2+4ϵ+4$$$$\iff \frac{{ϵ}^2-4ϵ+4}{2}<\mathrm{x}<\frac{{ϵ}^2+4ϵ+4}{2}$$$$\iff \frac{{ϵ}^2-4ϵ}{2}<\mathrm{x}-2<\frac{{ϵ}^2+4ϵ}{2}.\mathrm{Then}\mathrm{choosing}$$$$\delta =\frac{4ϵ-{ϵ}^2}{2}\mathrm{we}\mathrm{need}\mathrm{prove}\mathrm{that}\mathrm{\forall }\mathrm{x}\mathrm{so}\mathrm{that}$$$$\mid \mathrm{x}-2\mid <\delta =\frac{4ϵ-{ϵ}^2}{2}\mathrm{then}\mid \sqrt{2\mathrm{x}}-2\mid <ϵ.\mathrm{Ineed},$$$$\mid \mathrm{x}-2\mid <\delta =\frac{4ϵ-{ϵ}^2}{2}\iff -\delta +2<\mathrm{x}<2+\delta$$$$\iff -2\delta +4<2\mathrm{x}<4+2\delta$$$$\Rightarrow {ϵ}^2-4ϵ+4<2\mathrm{x}<4ϵ-{ϵ}^2+4<4+4ϵ+{ϵ}^2$$$$\Rightarrow 2-ϵ<\sqrt{2\mathrm{x}}<ϵ+2\Rightarrow \mid \sqrt{2\mathrm{x}-2}\mid <ϵ$$$$\mathrm{That}\mathrm{shows}\underset{\mathrm{x}\to 2}{\lim }\sqrt{2\mathrm{x}}=2(\mathbf{Q}.\mathbf{E}.\mathbf{D})$$

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