A sequence of real numbers $x_1, x_2, x_3, \dots$ satisfies the recurrence relation $x_{n+1} = \frac{5x_n + 2}{2x_n + 3} \text{ for all } n \ge 1.$

1. Set up the equation for the limit $l$ by applying the convergence property.

As $n \to \infty$, $x_n \to l$ and $x_{n+1} \to l$.
$l = \frac{5l + 2}{2l + 3}$

2. Rearrange the equation into a standard quadratic form.

$l(2l + 3) = 5l + 2$
$2l^2 + 3l = 5l + 2$
$2l^2 - 2l - 2 = 0$
$l^2 - l - 1 = 0$

3. Solve the quadratic equation for exact values of $l$.

$l = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$
$l = \frac{1 \pm \sqrt{1 + 4}}{2}$
$l = \frac{1 \pm \sqrt{5}}{2}$

1. Generate the first few terms of the sequence using the given recurrence relation and $x_1 = 3$.

Using a calculator:
$x_1 = 3$
$x_2 = \frac{5(3) + 2}{2(3) + 3} = \frac{17}{9} \approx 1.889$
$x_3 = \frac{5(17/9) + 2}{2(17/9) + 3} = \frac{103}{61} \approx 1.689$
$x_4 = \frac{5(103/61) + 2}{2(103/61) + 3} = \frac{637}{389} \approx 1.638$

2. Determine the behaviour of the sequence based on the calculated terms.

Observing the values, $x_1 > x_2 > x_3 > x_4$. The terms are strictly decreasing and approaching approximately $1.618$, which corresponds to the exact limit $l = \frac{1+\sqrt{5}}{2}$ found from solving $l = \frac{5l + 2}{2l + 3}$ in part (a).
Therefore, the sequence decreases and converges to $\frac{1+\sqrt{5}}{2}$.

1. Rearrange the given recurrence relation to make $x_n$ the subject.

$x_{n+1} = \frac{5x_n + 2}{2x_n + 3}$
$x_{n+1}(2x_n + 3) = 5x_n + 2$
$2x_n x_{n+1} + 3x_{n+1} = 5x_n + 2$
$2x_n x_{n+1} - 5x_n = 2 - 3x_{n+1}$
$x_n(2x_{n+1} - 5) = 2 - 3x_{n+1}$
$x_n = \frac{2 - 3x_{n+1}}{2x_{n+1} - 5}$

2. Use the inverse recurrence relation iteratively to compute $x_4, x_3, x_2$, and $x_1$ backwards from $x_5$.

Given $x_5 = \frac{3503}{2158}$, using a graphing calculator to compute the previous terms iteratively:
$x_4 = \frac{2 - 3\left(\frac{3503}{2158}\right)}{2\left(\frac{3503}{2158}\right) - 5} = \frac{563}{344}$
$x_3 = \frac{2 - 3\left(\frac{563}{344}\right)}{2\left(\frac{563}{344}\right) - 5} = \frac{91}{54}$
$x_2 = \frac{2 - 3\left(\frac{91}{54}\right)}{2\left(\frac{91}{54}\right) - 5} = \frac{15}{8}$
$x_1 = \frac{2 - 3\left(\frac{15}{8}\right)}{2\left(\frac{15}{8}\right) - 5} = \frac{29}{10}$