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,
$x = \dfrac{-1+\sqrt {13}}{2}$