Question Number 163137 by mathocean1 last updated on 04/Jan/22
$${Soit}\:{U}_{{n}} =\frac{{n}−\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{5}}\:{avec}\:{n}\:\in\:\mathbb{N}. \\ $$$${Montrer}\:{par}\:{la}\:{definition}\:{que}\:{U}_{{n}} \: \\ $$$${converge}\:{vers}\:\mathrm{0}. \\ $$
Answered by puissant last updated on 04/Jan/22
$${Soit}\:\varepsilon>\mathrm{0},\:{cherchons}\:{N}_{\varepsilon} \in\mathbb{N}\:/\:\forall{n}\in\mathbb{N}, \\ $$$${n}>{N}_{\varepsilon} \:\:\Rightarrow\:\mid{U}_{{n}} \mid<\varepsilon. \\ $$$${en}\:{effet},\:{on}\:{a}: \\ $$$$\mid{U}_{{n}} \mid=\mid\frac{{n}−\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{5}}\mid\geqslant\mid\frac{{n}}{{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{5}}\mid\geqslant\mid\frac{{n}}{{n}^{\mathrm{2}} }\mid \\ $$$${n}>{N}_{\varepsilon} \:\Rightarrow\:\frac{\mathrm{1}}{{n}}<\varepsilon\:\Rightarrow\:{n}>\frac{\mathrm{1}}{\varepsilon}. \\ $$$${prendre}\:{N}_{\varepsilon} =\:{E}\left(\frac{\mathrm{1}}{\varepsilon}\right)+\mathrm{1}. \\ $$
Commented by mathocean1 last updated on 10/Jan/22
$${thanks}. \\ $$