All you're ever gonna be is "mean"

$\begin{aligned} x_n & = \sqrt{x_{n-1}x_{n-2}} \\ \log x_n & = \frac 12 \log x_{n-1} + \frac 12 \log x_{n-2} & \small \color{#3D99F6}{\text{Let }y_n = \log x_n} \\ y_n & = \frac 12 y_{n-1} + \frac 12 y_{n-2} \end{aligned}$

The characteristic equation is as follows:

$\begin{aligned} r^2 - \frac 12 r - \frac 12 & = 0 \\ 2r^2 - r - 1 & = 0 \\ (2r+1)(r-1) & = 0 \\ \implies r & = -\frac 12, \ 1 \end{aligned}$

\begin{aligned} \implies y_n & = a_1 \left(\frac {-1}2\right)^n + a_2 \\ a_1 \left(\frac {-1}2\right)^n + a_2 & = \log x_n \\ a_1 \left(\frac {-1}2\right)^\color{#D61F06}{0} + a_2 & = \log x_\color{#D61F06}{0} \\ a_1+a_2 & = \log 1 = 0 \\ \implies a_2 & = - a_1 \\ \color{#3D99F6}{a_1} \left(\frac {-1}2\right)^\color{#D61F06}{1} \color{#3D99F6}{- a_1} & = \log x_\color{#D61F06}{1} \\ - \frac 32 a_1 & = \log 1000000 = 6 \\ \implies a_1 & = -4 \end{aligned}

$\begin{aligned} \implies \log x_n & = 4 \left(1-\left(\frac {-1}2\right)^n\right) \\ x_n & = 10^{4 \left(1-\left(\frac {-1}2\right)^n\right)} \\ \implies \lim_{n \to \infty} x_n & = \lim_{n \to \infty} 10^{4 \left(1-\left(\frac {-1}2\right)^n\right)} \\ & = 10^4 = \boxed{10000} \end{aligned}$

$x_0=10^0$ and $x_1=10^6$ . Working our way in powers it is equivalent to say that $A_0=0$ , $A_1=6$ , and $2A_n=A_{n-1}+A_{n-2}$ for $n\geq2$ . The recurrence relation is solved by standard technique to get $A_\infty = \frac {A_0+2A_1}{3}\rightarrow A_\infty = 4\rightarrow x=10000.$