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Question Number 114710 by mohammad17 last updated on 20/Sep/20

Answered by 1549442205PVT last updated on 20/Sep/20

We need find δ>0 such that∀x,y :  ∣x−0∣<δ,∣y−0∣<δ(1) then  ∣x^2 +y^2 −0∣<8 (2).Because  It is clear when ∣x∣<1 and ∣y∣<1  then ∣x^2 +y^2 ∣=2<8 hence,choosing δ=1  then for ∀x,y satisfy (1) the inequality  (2)is alway true.Therefore, in the  available answers  we choose a  (the other options don′t satisfy)

$$\mathrm{We}\:\mathrm{need}\:\mathrm{find}\:\delta>\mathrm{0}\:\mathrm{such}\:\mathrm{that}\forall\mathrm{x},\mathrm{y}\:: \\ $$$$\mid\mathrm{x}−\mathrm{0}\mid<\delta,\mid\mathrm{y}−\mathrm{0}\mid<\delta\left(\mathrm{1}\right)\:\mathrm{then} \\ $$$$\mid\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{0}\mid<\mathrm{8}\:\left(\mathrm{2}\right).\mathrm{Because} \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{clear}\:\mathrm{when}\:\mid\mathrm{x}\mid<\mathrm{1}\:\mathrm{and}\:\mid\mathrm{y}\mid<\mathrm{1} \\ $$$$\mathrm{then}\:\mid\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \mid=\mathrm{2}<\mathrm{8}\:\mathrm{hence},\mathrm{choosing}\:\delta=\mathrm{1} \\ $$$$\mathrm{then}\:\mathrm{for}\:\forall\mathrm{x},\mathrm{y}\:\mathrm{satisfy}\:\left(\mathrm{1}\right)\:\mathrm{the}\:\mathrm{inequality} \\ $$$$\left(\mathrm{2}\right)\mathrm{is}\:\mathrm{alway}\:\mathrm{true}.\mathrm{Therefore},\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{available}\:\mathrm{answers}\:\:\mathrm{we}\:\mathrm{choose}\:\mathrm{a} \\ $$$$\left(\mathrm{the}\:\mathrm{other}\:\mathrm{options}\:\mathrm{don}'\mathrm{t}\:\mathrm{satisfy}\right) \\ $$

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