A sequence of numbers, $x_{n}$, satisfy the relation $x_{n+1}=\tfrac{1}{3}\left(2x_{n}+\dfrac{1}{x_{n}^{2}}\right),\quad n\in\mathbb{N}^{+}\cup\{0\},\quad 0<x_{0}<1.$

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

As $n \to \infty$, $x_n \to L$ and $x_{n+1} \to L$.
Substituting $L$ into the recurrence relation:
$L = \frac{1}{3}\left(2L + \frac{1}{L^2}\right)$

2. Solve the algebraic equation for $L$.

$3L = 2L + \frac{1}{L^2}$
$L = \frac{1}{L^2}$
$L^3 = 1$
$L = 1 \quad \text{(since } L \text{ is a real number)}$

1. Determine the limit $L$ from part (i) as it is required to use the condition $x_n > L$.

As $n \to \infty$, both $x_n \to L$ and $x_{n+1} \to L$.
$L = \frac{1}{3}\left(2L + \frac{1}{L^2}\right)$
$3L = 2L + \frac{1}{L^2}$
$L = \frac{1}{L^2}$
$L^3 = 1 \implies L = 1$

2. Express the difference $x_{n+1} - x_n$ as a single rational fraction.

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

3. Determine the signs of the numerator and denominator using the condition $x_n > L$.

Given that $x_n > L$ and from our result $L = 1$, we have $x_n > 1$.
For the numerator: Since $x_n > 1$, it follows that $x_n^3 > 1$, which means $1 - x_n^3 < 0$.
For the denominator: Since $x_n > 1$, $3x_n^2 > 0$.

4. Conclude the proof by showing the difference is strictly negative.

Since the numerator is strictly negative and the denominator is strictly positive,
$x_{n+1} - x_n = \frac{1 - x_n^3}{3x_n^2} < 0$
$\therefore x_{n+1} < x_n \text{ (shown)}$

1. Synthesize the findings from parts (i) and (ii) to describe the sequence.

From part (i), the sequence converges to $L = 1$.
From part (ii), $x_{n+1} < x_n$ for $n \ge 1$, which means the sequence is strictly decreasing for $n \ge 1$.
Therefore, the sequence decreases and converges to $1$.

1. Find the limit $L$ of the sequence from part (i) to relate $x_n$ and $d_n$.

From (i), at the limit $L$, $x_{n} \to L$ and $x_{n+1} \to L$.
$L = \frac{1}{3}\left(2L + \frac{1}{L^2}\right)$
$3L^3 = 2L^3 + 1$
$L^3 = 1 \implies L = 1$

Since $d_n = x_n - L$, we have $x_n = d_n + 1$ and $x_{n+1} = d_{n+1} + 1$.

2. Substitute $x_n$ and $x_{n+1}$ into the recurrence relation to express $d_{n+1}$ in terms of $d_n$.

$d_{n+1} + 1 = \frac{1}{3}\left(2(d_n + 1) + \frac{1}{(d_n + 1)^2}\right)$
$d_{n+1} = \frac{1}{3}\left(2d_n + 2 + (1 + d_n)^{-2}\right) - 1$

3. Expand $(1 + d_n)^{-2}$ using the standard series expansion up to the $d_n^2$ term.

Using the standard expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots$:
$(1 + d_n)^{-2} = 1 + (-2)d_n + \frac{(-2)(-3)}{2}d_n^2 + \dots$
$(1 + d_n)^{-2} \approx 1 - 2d_n + 3d_n^2$

4. Substitute the expansion back and simplify to show the required relationship.

$d_{n+1} \approx \frac{1}{3}\left(2d_n + 2 + 1 - 2d_n + 3d_n^2\right) - 1$
$d_{n+1} \approx \frac{1}{3}\left(3 + 3d_n^2\right) - 1$
$d_{n+1} \approx 1 + d_n^2 - 1$
$d_{n+1} \approx d_n^2 \quad \text{(shown)}$

5. State the range of values for which the approximation (binomial series expansion) is valid.

For the binomial expansion of $(1 + d_n)^{-2}$ to be valid and converge, we require:
$|d_n| < 1$
$-1 < d_n < 1$