We are given $v_1=2$ and $v_{n+1}=\frac{1}{2}\left(v_n+\frac{8}{v_n}\right)$.
First notice that all terms are positive. Since $v_1=2>0$, and if $v_n>0$, then $\frac{8}{v_n}>0$, so $v_{n+1}>0$.
Now suppose the sequence converges and let
$$\lim_{n\to\infty}v_n=L$$
Since all terms are positive, $L\geq0$. Also, the recurrence contains $\frac{8}{v_n}$, so the limit cannot be $0$. Thus $L>0$.
Taking limits on both sides of
$$v_{n+1}=\frac{1}{2}\left(v_n+\frac{8}{v_n}\right)$$
we get
$$L=\frac{1}{2}\left(L+\frac{8}{L}\right)$$
Multiply both sides by $2$:
$$2L=L+\frac{8}{L}$$
Subtract $L$ from both sides:
$$L=\frac{8}{L}$$
Multiply by $L$:
$$L^2=8$$
So,
$$L=\pm\sqrt8=\pm2\sqrt2$$
Since $L>0$, we take the positive value:
$$\boxed{\therefore L=2\sqrt2}$$
Answer: B.
