Is $\frac{\text{next maximal}}{\text{maximal}} <= 2$ true?

Let $G_n(X)$ denote the $n^{th}$ record largest gap between consecutive primes below $X$, which some call maximal gap.

Let $G_{n+1}(Y)$ denote the size of next larger maximal record gap in size from $G_n(X)$ with consecutive primes below Y.

For example, we know the forth maximal prime gap uses the primes 23 and 29. So, $G_4(23) = 29-23 = 6$. The next prime gap is 2, but the one following it is also 37-31 = 6. Yes, a gap of 2 is smaller than $29-23$, but $G_4(29) = 6$, and $G_4(31) = 6$ while not being a maximal. This holds until $G_5(89) = 8$ at the $5^{th}$ maximal of $97-89$. In other words, $G_n(X)$ is the worst case function for gaps until the next worst case, which happens at the next maximal.

Denote the primes $p_x$, $p_{x+1}$ and $p_y$, $p_{y+1}$ as the primes starting and ending the largest maximal gap less than $X$ and $Y$ respectfully.

Note: All gaps $g_i = p_{i+1} - p_i$, where $p_i < p_y$ are $g_i <= G_n(X)$. This includes $g_{y-1} = p_y - p_{y-1}$.

Can these statements be proved:

$$\frac{G_{n+1}(Y)}{G_n(X)} \le 2?$$

And,

$$\lim_{n \to \infty} \frac{G_{n+1}(Y)}{G_n(X)} = 1?$$

Here is my try on the first:

"The $n^{th}$ Ramanujan prime is the least integer $R_n$ for which $\pi(x) - \pi(x/2) \ge n$, for all $x \ge R_n$."
$2p_{i-n} > p_i$ for $i > k$ where $k = \pi(p_k) = \pi(R_n)$, i.e. $p_k$ is the $k^{th}$ prime and the $n^{th}$ Ramanujan prime.
Let $R_b$ be the largest Ramanujan prime $R_b < p_y \le Y$.
Let $p_i = p_y$, then the prime $p_{i+1} = p_{y+1}$ is bounded by $2p_{i-n+1} = 2p_{y-b+1}$ by the RPC. Under the assumption that the smaller gap is $g_{y-b} = G_n(p_{y-b})$, we see that the larger gap is $p_{y+1} - p_y = g_y = G_{n+1}(p_y) \le 2G_n(X)$.