一个正数序列 $x_1, x_2, x_3, \dots$ 满足递推关系
$x_{n+1} = \sqrt{10 - 3x_n} \text{, for } n = 1, 2, 3, \dots$

1. Set up the limit equation by substituting $l$ into the recurrence relation.

As $n \to \infty$, both $x_n \to l$ and $x_{n+1} \to l$.
$l = \sqrt{10 - 3l}$

2. Solve the resulting quadratic equation and reject the invalid root.

Squaring both sides:
$l^2 = 10 - 3l$
$l^2 + 3l - 10 = 0$
$(l + 5)(l - 2) = 0$
$l = -5 \text{ or } l = 2$

Since the sequence consists of positive numbers ($x_n > 0$ for all $n$), the limit must be $l \ge 0$. Also, the principal square root is non-negative. Therefore, $l = -5$ is rejected.
$l = 2$

1. Express $(x_{n+1})^2$ and $l^2$ in terms of $x_n$ and $l$ respectively.

From the recurrence relation $x_{n+1} = \sqrt{10 - 3x_n}$, squaring both sides gives:
$(x_{n+1})^2 = 10 - 3x_n$
Since the sequence converges to $l$, the limit $l$ satisfies the same relation:
$l = \sqrt{10 - 3l} \implies l^2 = 10 - 3l$

2. Evaluate the difference $(x_{n+1})^2 - l^2$ to prove the identity.

Subtracting the expression for $l^2$ from the expression for $(x_{n+1})^2$:
$(x_{n+1})^2 - l^2 = (10 - 3x_n) - (10 - 3l)$
$(x_{n+1})^2 - l^2 = 10 - 3x_n - 10 + 3l$
$(x_{n+1})^2 - l^2 = 3l - 3x_n$
$(x_{n+1})^2 - l^2 = 3(l - x_n)$

1. Establish the sign of the expression $3(l - x_n)$ given the inequality $x_n < l$.

Given $x_n < l$, we can rearrange the inequality to obtain:
$l - x_n > 0$
Multiplying by 3 gives:
$3(l - x_n) > 0$

2. Use the result from part (ii) to determine the relationship between $x_{n+1}$ and $l$.

From the result in part (ii), we are given:
$(x_{n+1})^2 - l^2 = 3(l - x_n)$
Since $3(l - x_n) > 0$, it follows that:
$(x_{n+1})^2 - l^2 > 0$
$(x_{n+1})^2 > l^2$
We are given that the sequence consists of positive numbers, so $x_{n+1} > 0$. From part (i), the limit is $l = 2 > 0$. Since both $x_{n+1}$ and $l$ are strictly positive, taking the square root of both sides preserves the inequality:
$x_{n+1} > l \quad \text{(Shown)}$