Solution: The given sequence is $$x_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}},n\ge 1.$$ Let us suppose that $x_0>0$, which implies $x_n>0$ for all $n\ge 0$. Now apply AM-GM inequality on $\frac{a_{n-1}}{2}$ and $\frac{1}{x_{n-1}}$. $$\begin{array}{c}\frac{x_n}{2}=\frac{\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}}{2}\ge \sqrt{\frac{x_{n-1}}{2}\cdot \frac{1}{x_{n-1}}}⟹\frac{x_n}{2}\ge \sqrt{2}.\end{array}$$ Thus, the sequence $\{x_n\}$ is bounded below by $\sqrt{2}$. Also, $$\begin{array}{rl}{x}_n-x_{n-1}& =x_n-[\frac{x_n}{2}+\frac{1}{x_n}]\\ & =\frac{x_n}{2}-\frac{1}{x_n}=\frac{x_n^2-2}{2x_n}.\end{array}$$ Now, observe that $$\begin{array}{rl}{x}_n^2-2& ={(\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}})}^2-2\\ & =\frac{x_{n-1}^2}{4}+\frac{1}{x_{n-1}^2}+1-2\\ & =\frac{x_{n-1}^2}{4}+\frac{1}{x_{n-1}^2}-1\\ & ={(\frac{x_{n-1}}{2}-\frac{1}{x_{n-1}})}^2\ge 0.\end{array}$$ Therefore, $$x_n-x_{n-1}\ge 0⟹x_n\ge x_{n-1}.$$ 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, $$\begin{array}{rl}{x}_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}& ⟹\underset{n\to \mathrm{\infty }}{lim}{x}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}\\ & ⟹\ell =\frac{\ell }{2}+\frac{1}{\ell }\\ & ⟹\frac{\ell }{2}=\frac{1}{2}\\ & ⟹{\ell }^2-2=0\\ & ⟹\ell =\pm \sqrt{2}.\end{array}$$ 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 $x_n\to -\sqrt{2}$. Thus, option (b) is the correct option.