Solution: The given sequence is \[ x_n = \frac{x_{n-1} }{2} + \frac{1}{x_{n-1} }, \quad n\geq 1. \] Let us suppose that $x_0 > 0$, which implies $x_n > 0$ for all $n \geq 0$. Now apply AM-GM inequality on $\frac{a_{n-1} }{2}$ and $\frac{1}{x_{n-1} }$. \begin{gather*} \frac{x_n}{2} = \frac{\frac{x_{n-1} }{2} + \frac{1}{ x_{n-1} }}{2} \geq \sqrt{\frac{x_{n-1} }{2} \cdot \frac{1}{x_{n-1}}} \implies \frac{x_n}{2} \geq \sqrt{2}. \end{gather*} Thus, the sequence $\{ x_n \} $ is bounded below by $\sqrt{2} $. Also, \begin{align*} x_n - x_{n-1} & = x_n - \left[ \frac{x_n}{2} + \frac{1}{x_n} \right] \\[1ex] & = \frac{x_n}{2} - \frac{1}{x_n} = \frac{x_n^2 - 2}{2 x_n}. \end{align*} Now, observe that \begin{align*} x_n^2 - 2 & = \left( \frac{x_{n-1} }{2} + \frac{1}{x_{n-1} } \right) ^2 - 2\\ & = \frac{x_{n-1}^2 }{4} + \frac{1}{x_{n-1}^2 } + 1 - 2\\ & = \frac{x_{n-1}^2}{4} + \frac{1}{x_{n-1}^2} - 1 \\ & = \left( \frac{x_{n-1} }{2} - \frac{1}{x_{n-1} } \right) ^2 \geq 0. \end{align*} Therefore, \[ x_n - x_{n-1} \geq 0 \implies x_n \geq x_{n-1}. \] Thus proves that the sequence $\{ x_n \} $ is monotonically decreasing sequence. Hence, by the monotone convergence theorem, the sequence $\{ x_n \} $ is convergent, let say it converges to $\ell$.

As $x_n \to \ell$ so $x_{n-1} \to \ell$. Thus, \begin{align*} x_n = \frac{x_{n-1} }{2} + \frac{1}{x_{n-1} } & \implies \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{x_{n-1} }{2} + \frac{1}{x_{n-1} } \\ & \implies \ell = \frac{\ell}{2} + \frac{1}{\ell} \\ & \implies \frac{\ell}{2} = \frac{1}{2} \\ & \implies \ell^2 - 2 = 0 \\ & \implies \ell = \pm \sqrt{2}. \end{align*} Since, all the terms of the sequence is positive, the limit must be nonnegative. Hence, $l = \sqrt{2} $.

imilarly, if $x_0 < 0$, then one can show that $x_n \rightarrow -\sqrt{2} $. Thus, option (b) is the correct option.