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Question Number 145391 by puissant last updated on 04/Jul/21

$$\mathrm{Developpement}\mathrm{limit}\acute{\mathrm{e}}\mathrm{a}{\mathrm{l}}^{\prime }\mathrm{ordre}2\mathrm{de}$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\sqrt{1+{\mathrm{x}}^2}}{1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}$$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

$$g\left(x\right)=\frac{\sqrt{1+x^2}}{1+x+\sqrt{1+x^2}}$$$$g\left(x\right)=\frac{\sqrt{1+x^2}}{{(1+x)}^2-(1+x^2)}(1+x-\sqrt{1+x^2})$$$$g\left(x\right)=\frac{\sqrt{1+x^2}}{2x}(1+x-\sqrt{1+x^2})$$$$g\left(x\right)=\frac{1+\frac{x^2}{2}}{2x}(1+x-1-\frac{x^2}{2})+o\left(x^2\right)$$$$g\left(x\right)=\frac{1+\frac{x^2}{2}}{2}(1-\frac{x}{2})+o\left(x^2\right)$$$$g\left(x\right)=\frac{1}{2}(1+\frac{x^2}{2}-\frac{x}{2})+o\left(x^2\right)$$$$g\left(x\right)=\frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}+o\left(x^2\right)$$

Commented by puissant last updated on 04/Jul/21

$$\mathrm{merci}$$

Answered by mathmax by abdo last updated on 04/Jul/21

$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{g}\left(0\right)+{\mathrm{xg}}^{{}^{\prime }}\left(0\right)+\frac{{\mathrm{x}}^2}{2}{\mathrm{g}}^{\left(2\right)}\left(0\right)+{\mathrm{x}}^3\xi \left(\mathrm{x}\right)$$$$\mathrm{g}\left(0\right)=\frac{1}{2}\mathrm{we}\mathrm{have}{\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{\frac{2\mathrm{x}}{2\sqrt{1+{\mathrm{x}}^2}}(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-(1+\frac{2\mathrm{x}}{2\sqrt{1+{\mathrm{x}}^2}})\sqrt{1+{\mathrm{x}}^2}}{{(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}^2}$$$$=\frac{\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\left(\frac{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}{\sqrt{1+{\mathrm{x}}^2}}\right)\sqrt{1+{\mathrm{x}}^2}}{{(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}^2}$$$$=\frac{(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\sqrt{1+{\mathrm{x}}^2}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}{\sqrt{1+{\mathrm{x}}^2}{(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}^2}\Rightarrow$$$${\mathrm{g}}^{{}^{\prime }}\left(0\right)=\frac{2-1}{2}=\frac{1}{2}$$$${\mathrm{g}}^{\left(2\right)}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}-\mathrm{v}}{\mathrm{w}}\right)=\frac{({\mathrm{u}}^{{}^{\prime }}-{\mathrm{v}}^{{}^{\prime }})\mathrm{w}-(\mathrm{u}-\mathrm{v}){\mathrm{w}}^{{}^{\prime }}}{{\mathrm{w}}^2}$$$$\mathrm{u}=1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\Rightarrow {\mathrm{u}}^{{}^{\prime }}=1+\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}$$$$\mathrm{v}=\sqrt{1+{\mathrm{x}}^2}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})\Rightarrow {\mathrm{v}}^{{}^{\prime }}=\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})+\sqrt{1+{\mathrm{x}}^2}(1+\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}})$$$$\mathrm{w}=\sqrt{1+{\mathrm{x}}^2}{(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}^2\Rightarrow$$$${\mathrm{w}}^{{}^{\prime }}=\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}{(1+\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}^2+2\sqrt{1+{\mathrm{x}}^2}(1+\mathrm{x}+\sqrt{1+x^2})(1+\frac{x}{\sqrt{1+x^2}})\Rightarrow$$$${\mathrm{g}}^{\left(2\right)}\left(0\right)=\frac{({\mathrm{u}}^{{}^{\prime }}\left(0\right)-{\mathrm{v}}^{{}^{\prime }}\left(0\right))\mathrm{w}\left(0\right)-(\mathrm{u}\left(0\right)-\mathrm{v}\left(0\right)){\mathrm{w}}^{{}^{\prime }}\left(0\right)}{{\mathrm{w}}^2\left(0\right)}$$$$=\frac{(1-1)4-(2-1)4}{4^2}=-\frac{1}{4}\Rightarrow$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{1}{2}+\frac{\mathrm{x}}{2}-\frac{{\mathrm{x}}^2}{8}+{\mathrm{x}}^3\xi \left(\mathrm{x}\right)$$

Commented by mathmax by abdo last updated on 04/Jul/21

$$\mathrm{sorry}{\mathrm{g}}^{{}^{\prime }}\left(0\right)=\frac{1}{4}\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\frac{1}{2}+\frac{\mathrm{x}}{4}-\frac{{\mathrm{x}}^2}{8}+{\mathrm{x}}^3\xi \left(\mathrm{x}\right)$$

Commented by puissant last updated on 04/Jul/21

$$\mathrm{thanks}$$

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