bonsoir j ei besoin d'aide pour resoudre cet exercice 1/calculer la limite

[tex]Bonsoir;\\\\\\ \textit{On a : } \dfrac{ \sqrt{2x-1} -\sqrt{x-1} - 1 }{x-1} = \dfrac{ \sqrt{2x-1} - 1 }{x-1} - \dfrac{ \sqrt{x-1} }{x-1} \\\\\\ = \dfrac{ (\sqrt{2x-1} - 1)(\sqrt{2x-1} + 1) }{(x-1)(\sqrt{2x-1} + 1)} - \dfrac{ \sqrt{x-1} }{\sqrt{x-1}^2} \\\\\\ = \dfrac{ \sqrt{2x-1}^2 - 1 }{(x-1)(\sqrt{2x-1} + 1)} - \dfrac{ 1 }{\sqrt{x-1}} = \dfrac{ 2x-1 - 1 }{(x-1)(\sqrt{2x-1} + 1)} - \dfrac{ 1 }{\sqrt{x-1}} \\\\\\ = \dfrac{ 2(x-1 ) }{(x-1)(\sqrt{2x-1} + 1)} - \dfrac{ 1 }{\sqrt{x-1}} [/tex]


[tex]= \dfrac{ 2}{\sqrt{2x-1} + 1} - \dfrac{ 1 }{\sqrt{x-1}} ; \\\\\\ \textit{comme on a :} \underset{x\rightarrow 1^+}{lim} \dfrac{ 2}{\sqrt{2x-1} + 1} = 1 \textit{ et } \underset{x\rightarrow 1^+}{lim} \dfrac{ 1 }{\sqrt{x-1}} = +\infty \ ;\\\\\\ \textit{alors : } \underset{x\rightarrow 1^+}{lim} \dfrac{ 2}{\sqrt{2x-1} + 1} - \dfrac{ 1 }{\sqrt{x-1}} = - \infty \ ; \\\\\\ \textit{alors : } \underset{x\rightarrow 1^+}{lim} \dfrac{ \sqrt{2x-1} -\sqrt{x-1} - 1 }{x-1} = - \infty \ .[/tex]