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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
$we have x_0 =1 and x_(n+1) =((3+2x_n )/(3+x_n )) ⇒x_(n+1) =f(x_n ) with f(x)=((2x+3)/(x+3)) f is continue on R−{−3} and lim_(x→+∞) =2 f^′ (x) =((2(x+3)−(2x+3)×1)/((x+3)^2 )) =((2x+6−2x−3)/((x+3)^2 )) =(3/((x+3)^2 )) >0 ⇒ f is increazing onR−{−3} let determine the fixed point f(x)=x ⇒((2x+3)/(x+3)) =x ⇒2x+3 =x^2 +3x ⇒x^2 +3x−2x−3=0 ⇒ x^2 +x−3 =0 →Δ =1−4(−3) =13 ⇒x_1 =((−1+(√(13)))/2) and x_2 =((−1−(√(13)))/2) but x_n >0 for all n ⇒lim_(n→+∞) x_n =(((√(13))−1)/2) .$
$w e h a v e x_{0} = 1 a n d x_{n + 1} = \frac{3 + 2 x_{n}}{3 + x_{n}} \Rightarrow x_{n + 1} = f (x_{n}) w i t h$ $f (x) = \frac{2 x + 3}{x + 3} f i s c o n t i n u e o n R - {- 3} a n d {l i m}_{x \to + \infty} = 2$ $f^{^{'}} (x) = \frac{2 (x + 3) - (2 x + 3) \times 1}{{(x + 3)}^{2}} = \frac{2 x + 6 - 2 x - 3}{{(x + 3)}^{2}} = \frac{3}{{(x + 3)}^{2}} > 0 \Rightarrow$ $f i s i n c r e a z i n g o n R - {- 3} l e t d e t e r m i n e t h e f i x e d p o i n t$ $f (x) = x \Rightarrow \frac{2 x + 3}{x + 3} = x \Rightarrow 2 x + 3 = x^{2} + 3 x \Rightarrow x^{2} + 3 x - 2 x - 3 = 0 \Rightarrow$ $x^{2} + x - 3 = 0 \to Δ = 1 - 4 (- 3) = 13 \Rightarrow x_{1} = \frac{- 1 + \sqrt{13}}{2}$ $a n d x_{2} = \frac{- 1 - \sqrt{13}}{2} b u t x_{n} > 0 f o r a l l n \Rightarrow {l i m}_{n \to + \infty} x_{n} = \frac{\sqrt{13} - 1}{2} .$
Commented by aliesam last updated on 12/Jul/19

$g o d b l e s s y o u s i r$
Commented by mathmax by abdo last updated on 12/Jul/19

$y o u a r e w e l c o m e .$
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