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Question Number 158341 by aliibrahim1 last updated on 02/Nov/21

Answered by puissant last updated on 03/Nov/21

$$Thatis\epsilon >0,{let}^{\prime }sseek\alpha >0/\mathrm{\forall }x\in \mathbb{R},0<\mid x-2\mid <\alpha \Rightarrow \mid x^2-4\mid <\epsilon$$$$forx\in \mathbb{R},\mid x^2-4\mid =\mid x-2\mid .\mid x+2\mid wehave\mid x-2\mid <2\to 0<x+2<6$$$$\Rightarrow \mid x^2-4\mid <6\mid x-2\mid ;\mid x^2-4\mid <\epsilon \to 6\mid x-2\mid <\epsilon \to \mid x-2\mid <\frac{\epsilon }{6}$$$$take\alpha =min(2;\frac{\epsilon }{6})....◼$$$$$$$$.................\mathcal{L}epuissant...............$$

Commented by aliibrahim1 last updated on 04/Nov/21

$$thxsir$$

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