Question Number 4728 by 123456 last updated on 01/Mar/16
$$\mathrm{if}\:\mid{x}^{\mathrm{2}} −{c}\mid\leqslant\epsilon,\:{c}\in\mathbb{R},{c}\geqslant\mathrm{0} \\ $$$$\mathrm{does}? \\ $$$$\mid{x}−\sqrt{{c}}\mid\leqslant\epsilon \\ $$
Commented by Yozzii last updated on 01/Mar/16
$$\mid\left({x}−\sqrt{{c}}\right)\mid\mid{x}+\sqrt{{c}}\mid\leqslant\epsilon \\ $$$$\mid{x}−\sqrt{{c}}\mid\leqslant\frac{\epsilon}{\mid{x}+\sqrt{{c}}\mid}\:\left({x}\neq−\sqrt{{c}}\right) \\ $$$${If}\:\mid{x}+\sqrt{{c}}\mid\geqslant\mathrm{1}\Rightarrow\mathrm{1}\geqslant\frac{\mathrm{1}}{\mid{x}+\sqrt{{c}}\mid} \\ $$$$\Rightarrow\epsilon\geqslant\frac{\epsilon}{\mid{x}+\sqrt{{c}}\mid}\Rightarrow\mid{x}−\sqrt{{c}}\mid\leqslant\epsilon. \\ $$$${It}\:{appears}\:{to}\:{be}\:{necessary}\:{that}\: \\ $$$${x}\geqslant\mathrm{1}−\sqrt{{c}}\:{or}\:{x}\leqslant−\mathrm{1}−\sqrt{{c}}\:{to}\:{deduce}\: \\ $$$$\mid{x}−\sqrt{{c}}\mid\leqslant\epsilon\:\:\:\:\left(\epsilon>\mathrm{0}\right). \\ $$$${Suppose}\:{x}=\mathrm{1}−\sqrt{\mathrm{25}}−\mathrm{1}=−\mathrm{5}\: \\ $$$${for}\:{c}=\mathrm{25}\:\left({so}\mid{x}−\sqrt{{c}}\mid<\mathrm{1}\right). \\ $$$$\Rightarrow\mid\mathrm{25}−\mathrm{25}\mid=\mathrm{0}\leqslant\mathrm{1}\:\left({so}\:{say}\:\epsilon=\mathrm{1}\right). \\ $$$${But}\mid{x}−\sqrt{{c}}\mid=\mid−\mathrm{5}−\mathrm{5}\mid=\mathrm{10}\nleqq\mathrm{1}=\epsilon. \\ $$$${So},\:\left({x},{c},\epsilon\right)=\left(−\mathrm{5},\mathrm{25},\mathrm{1}\right)\:{shows}\:{the} \\ $$$${original}\:{implication}\:{being}\:{incorrect}. \\ $$