Limit Problem Of Different Number System

Note that the number of digits of $n$ in base $b$ is given by $d_b (n) = \lfloor \log_b n \rfloor + 1$ , where $n$ is in decimal.

$\begin{aligned} L & = \lim_{n \to \infty} \frac {p(n)q(n)}{n^2} \\ & = \lim_{n \to \infty} \frac {\left(\left \lfloor \log_6 8^n \right \rfloor + 1\right)\left(\left \lfloor \log_4 6^n \right \rfloor + 1\right)}{n^2} \\ & = \lim_{n \to \infty} \frac {\left \lfloor n \log_6 8 \right \rfloor \left \lfloor n \log_4 6 \right \rfloor + \left \lfloor n \log_6 8 \right \rfloor + \left \lfloor n \log_4 6 \right \rfloor + 1}{n^2} & \small \color{#3D99F6} \text{For very large }n, \\ & = \lim_{n \to \infty} \frac {n^2 \log_6 8 \log_4 6 + n \log_6 8 + n \log_4 6 + 1}{n^2} & \small \color{#3D99F6} \text{Divide up and down by }n^2 \\ & = \lim_{n \to \infty} \frac {\log_6 8 \log_4 6 + \frac {\log_6 8}n + \frac {\log_4 6}n + \frac 1{n^2}}1 \\ & = \log_6 8 \log_4 6 = \frac {\log_2 8}{\log_2 6} \times \frac {\log_2 6}{\log_2 4} = \frac 32 = \boxed{1.5} \end{aligned}$