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Let $x_{0}=1$ and

$x_{n+1}= \frac{3+2x_{n}}{3+x_{n}}, n\geq 0$.

$x_{\infty}=\displaystyle \lim_{n\rightarrow \infty}x_{n}$ is

  1. $\left(\sqrt{5}-1\right) / 2$
  2. $\left(\sqrt{5}+1\right) / 2$
  3. $\left(\sqrt{13}-1\right) / 2$
  4. $\left(-\sqrt{13}-1\right) / 2$
  5. None of the above
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Answer : C

$x_{n+1}=1+\dfrac{x(n)}{3+x(n)}$

As $n$ tends to infinity, $x_{n+1} = x_n = x$

$x=1+\dfrac{x}{3+x}$

$\implies x^2+3x=3+2x$

$\implies x^2 + x - 3 = 0$

The roots are: $\dfrac{-1+ \sqrt{13}} {2}, \dfrac{-1 - \sqrt{13}} {2}$

Since, $x_n$ is positive,

$x = \dfrac{-1+\sqrt {13}}{2}$
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clearly x(i) >= 1; now dividing numerator and denominator of x(n+1) by x(n)

x(n+1)= (3(/x(n))+2 )/(3/x(n)+1)

but as n increases, 1/(x(n) approaches 0;

therefore limit n-> infinity, x(n+1)=(0+2)/(0+1)=2.
Answer:

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