一个负实数数列 $x_{1}, x_{2}, x_{3}, \dots$ 满足关系式 $x_{n+1} = -\sqrt{1-2x_{n}}, \text{ for } n\ge 1.$ 已知该数列收敛于 $l$，求 $l$ 的精确值。

1. Set up the equation for the limit $l$.

Since the sequence converges to $l$, as $n \to \infty$, both $x_{n} \to l$ and $x_{n+1} \to l$. Substituting this into the recurrence relation gives:
$l = -\sqrt{1-2l}$

2. Square both sides and solve the resulting quadratic equation.

$l^2 = 1 - 2l$
$l^2 + 2l - 1 = 0$
Using the quadratic formula:
$l = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2}$
$l = \frac{-2 \pm \sqrt{8}}{2}$
$l = \frac{-2 \pm 2\sqrt{2}}{2}$
$l = -1 \pm \sqrt{2}$

3. Determine the correct value of $l$ by checking for extraneous roots.

Since the sequence consists of negative real numbers, or from observing the equation $l = -\sqrt{1-2l}$ where the principal square root is non-negative, it must be that $l \le 0$.
Since $\sqrt{2} \approx 1.414$, we have $-1 + \sqrt{2} > 0$.
Thus, $l = -1 + \sqrt{2}$ is an extraneous root introduced by squaring and must be rejected.
$l = -1 - \sqrt{2}$