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Question Number 64047 by aliesam last updated on 12/Jul/19

Commented by mathmax by abdo last updated on 12/Jul/19

we have x_0 =1 and x_(n+1) =((3+2x_n )/(3+x_n )) ⇒x_(n+1) =f(x_n ) with  f(x)=((2x+3)/(x+3))   f is continue on R−{−3} and lim_(x→+∞) =2  f^′ (x) =((2(x+3)−(2x+3)×1)/((x+3)^2 )) =((2x+6−2x−3)/((x+3)^2 )) =(3/((x+3)^2 )) >0 ⇒  f is increazing onR−{−3} let determine the fixed point  f(x)=x ⇒((2x+3)/(x+3)) =x ⇒2x+3 =x^2  +3x ⇒x^2  +3x−2x−3=0 ⇒  x^2  +x−3 =0 →Δ =1−4(−3) =13 ⇒x_1 =((−1+(√(13)))/2)  and x_2 =((−1−(√(13)))/2)   but  x_n >0  for all n  ⇒lim_(n→+∞)  x_n =(((√(13))−1)/2) .

$${we}\:{have}\:{x}_{\mathrm{0}} =\mathrm{1}\:{and}\:{x}_{{n}+\mathrm{1}} =\frac{\mathrm{3}+\mathrm{2}{x}_{{n}} }{\mathrm{3}+{x}_{{n}} }\:\Rightarrow{x}_{{n}+\mathrm{1}} ={f}\left({x}_{{n}} \right)\:{with} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{3}}{{x}+\mathrm{3}}\:\:\:{f}\:{is}\:{continue}\:{on}\:{R}−\left\{−\mathrm{3}\right\}\:{and}\:{lim}_{{x}\rightarrow+\infty} =\mathrm{2} \\ $$$${f}^{'} \left({x}\right)\:=\frac{\mathrm{2}\left({x}+\mathrm{3}\right)−\left(\mathrm{2}{x}+\mathrm{3}\right)×\mathrm{1}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}{x}+\mathrm{6}−\mathrm{2}{x}−\mathrm{3}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }\:>\mathrm{0}\:\Rightarrow \\ $$$${f}\:{is}\:{increazing}\:{onR}−\left\{−\mathrm{3}\right\}\:{let}\:{determine}\:{the}\:{fixed}\:{point} \\ $$$${f}\left({x}\right)={x}\:\Rightarrow\frac{\mathrm{2}{x}+\mathrm{3}}{{x}+\mathrm{3}}\:={x}\:\Rightarrow\mathrm{2}{x}+\mathrm{3}\:={x}^{\mathrm{2}} \:+\mathrm{3}{x}\:\Rightarrow{x}^{\mathrm{2}} \:+\mathrm{3}{x}−\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+{x}−\mathrm{3}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}\left(−\mathrm{3}\right)\:=\mathrm{13}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${and}\:{x}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{2}}\:\:\:{but}\:\:{x}_{{n}} >\mathrm{0}\:\:{for}\:{all}\:{n}\:\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{x}_{{n}} =\frac{\sqrt{\mathrm{13}}−\mathrm{1}}{\mathrm{2}}\:. \\ $$

Commented by aliesam last updated on 12/Jul/19

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 12/Jul/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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