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Question Number 190162 by yaslm last updated on 29/Mar/23

Answered by Skabetix last updated on 29/Mar/23

0^2 +2^2 +3=4+3=7

02+22+3=4+3=7

Commented by yaslm last updated on 29/Mar/23

prove use delta and epsilon

Answered by mehdee42 last updated on 29/Mar/23

we have to show ∀ε>0∃δ>0 ; −∥(x,y)−(0,2∥<δ⇒∣x^2 +y^2 +3−7∣<ε if δ_1 =1⇒∣x∣<1 & 2<y<3⇒∣y+2∣<5 ∣x^2 +y^2 −4∣=∣x^2 +(y+2)(y−2)∣≤∣x∣∣x∣+∣y+2∣∣y−2∣ ≤∣x∣(√(x^2 +(y−2)^2 ))+∣y+2∣(√(x^2 +(y−2)^2 ))< <(√(x^2 +(y−2)^2 ))+5(√(x^2 +(y−2)^2 ))<6(√(x^2 +(y−2)^2 ))=6∥(x,y)−(0,2)∥<ε ⇒we choose δ=min{δ_1 ,(ε/6)}

wehavetoshow∀ε>0∃δ>0;−∥(x,y)−(0,2∥<δ⇒∣x2+y2+3−7∣<εifδ1=1⇒∣x∣<1&2<y<3⇒∣y+2∣<5∣x2+y2−4∣=∣x2+(y+2)(y−2)∣⩽∣x∣∣x∣+∣y+2∣∣y−2∣⩽∣x∣x2+(y−2)2+∣y+2∣x2+(y−2)2<<x2+(y−2)2+5x2+(y−2)2<6x2+(y−2)2=6∥(x,y)−(0,2)∥<ε⇒wechooseδ=min{δ1,ε6}

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