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Question Number 55641 by gunawan last updated on 01/Mar/19

$$\mathrm{Studies}\mathrm{of}\mathrm{convergences}$$$$\mathrm{the}\mathrm{numbers}\mathrm{real}\mathrm{sequence}\left\{x_n\right\},$$$$\mathrm{with}{x}_1=1\mathrm{and}{x}_{n+1}=\frac{x_n^2+2}{2x_n},n⩾1$$$$$$$$$$

Commented by maxmathsup by imad last updated on 01/Mar/19

$$wehave{x}_{n+1}=f\left(x_n\right)withf\left(x\right)=\frac{x^2+2}{2x}=\frac{x}{2}+\frac{1}{x}withx>0$$$$we{havef}^{{}^{\prime }}\left(x\right)=\frac{1}{2}-\frac{1}{x^2}=\frac{x^2-2}{2x^2}and{f}^{{}^{\prime }}\left(x\right)=0\iff x=\stackrel{-}{+}\sqrt{2}$$$$x0\sqrt{2}+\mathrm{\infty }$$$$f^{{}^{\prime }}\left(x\right)-+$$$$f\left(x\right)+\mathrm{\infty }deccf\left(\sqrt{2}\right)incre+\mathrm{\infty }$$$$f\left(\sqrt{2}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$$$$fiscontinueon]0,+\mathrm{\infty }[letfindthefixpoint$$$$f\left(x\right)=x\iff \frac{x}{2}+\frac{1}{x}=x\iff \frac{1}{x}=\frac{x}{2}\Rightarrow 2=x^2\Rightarrow x=\sqrt{2}becausex>0$$$$\left(x_n\right)isconvergentand{lim}_{n\to +\mathrm{\infty }}{x}_n=\sqrt{2}.$$

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