Figure 8 shows a sketch of the curve C with equation $y = x^{rac{1}{3}}$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 13 - 2019 - Paper 2

Given the point $P(a, 2)$ lies on the curve, we substitute into the original equation:

$2 = a^{\frac{1}{3}} \Rightarrow a = 2^3 = 8.$

Now evaluating $x^{\frac{1}{3}}$ near the vicinity of $2$:
For $a = 1.5$,
$1.5^{\frac{1}{3}} \approx 1.144\ ext{ and for } a = 1.6, \\ 1.6^{\frac{1}{3}} \approx 1.171.$
This shows that as $a$ approaches between 1.5 and 1.6, the values of $y$ remain between $(1.144, 1.171)$, confirming that $1.5 < a < 1.6$.

Using the iteration formula $x_{n+1} = 2x_n^2$, we start with $x_1 = 1.5$:

1. Calculate $x_2$:
   $x_2 = 2(1.5)^2 = 2(2.25) = 4.5.$

2. Calculate $x_3$:
   $x_3 = 2(4.5)^2 = 2(20.25) = 40.5.$

Thus, $x_3 = 40.5$.

As we apply the iteration formula repeatedly, the sequence $x_n$ increases rapidly since with each iteration we square the previous value and multiply by 2. Therefore, as $n$ approaches infinity, $x_n$ heads towards infinity:

$\lim_{n \to \infty} x_n = \infty.$
Hence, the long-term behaviour of $x_n$ is that it diverges to infinity.