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Question Number 199175 by SANOGO last updated on 29/Oct/23

Answered by a.lgnaoui last updated on 29/Oct/23

$${\mathrm{g}}_{\mathrm{m}}\left(\mathrm{x}\right)={\mathrm{x}}^2+\frac{2\mathrm{x}}{\mathrm{m}}+\frac{1}{{\mathrm{m}}^2}(\mathrm{m}\ne 0\mathrm{x}⩾0)$$$$\mathrm{n}\mathrm{est}\mathrm{pas}\mathrm{une}\mathrm{suite}\mathrm{convergente}$$$$\mathrm{preuve}:$$$$\lim {\mathrm{g}}_{\mathrm{m}}{\left(\mathrm{x}\right)}_{\mathrm{x}\to +\mathrm{\infty }}={\mathrm{limx}}_{\mathrm{x}\to \mathrm{\infty }}{(\times +1)}^2$$$$={\mathrm{limg}}_1{\left(\mathrm{x}\right)}_{\mathrm{x}\to +\mathrm{\infty }}=+\mathrm{\infty }$$$${(\mathbf{x}+1)}^2{\mathbf{n}}^{\prime }\mathrm{est}\mathrm{pas}\mathrm{une}\mathrm{suute}\mathrm{de}\mathrm{Cauchy}$$$$\mathrm{donc}{\mathbf{g}}_{\mathbf{m}}\left(\mathbf{x}\right)\mathbf{est}\mathbf{une}\mathbf{suite}\mathbf{divergente}$$$$$$

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