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Question-190162




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
$$\mathrm{0}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}=\mathrm{4}+\mathrm{3}=\mathrm{7} \\ $$
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)}
$${we}\:{have}\:{to}\:{show}\: \\ $$$$\forall\varepsilon>\mathrm{0}\exists\delta>\mathrm{0}\:;\:−\parallel\left({x},{y}\right)−\left(\mathrm{0},\mathrm{2}\parallel<\delta\Rightarrow\mid{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}−\mathrm{7}\mid<\varepsilon\right. \\ $$$${if}\:\:\delta_{\mathrm{1}} =\mathrm{1}\Rightarrow\mid{x}\mid<\mathrm{1}\:\&\:\mathrm{2}<{y}<\mathrm{3}\Rightarrow\mid{y}+\mathrm{2}\mid<\mathrm{5} \\ $$$$\mid{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}\mid=\mid{x}^{\mathrm{2}} +\left({y}+\mathrm{2}\right)\left({y}−\mathrm{2}\right)\mid\leqslant\mid{x}\mid\mid{x}\mid+\mid{y}+\mathrm{2}\mid\mid{y}−\mathrm{2}\mid \\ $$$$\leqslant\mid{x}\mid\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }+\mid{y}+\mathrm{2}\mid\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }< \\ $$$$<\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }+\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }<\mathrm{6}\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{6}\parallel\left({x},{y}\right)−\left(\mathrm{0},\mathrm{2}\right)\parallel<\varepsilon \\ $$$$\Rightarrow{we}\:{choose}\:\:\delta={min}\left\{\delta_{\mathrm{1}} ,\frac{\varepsilon}{\mathrm{6}}\right\} \\ $$$$ \\ $$

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