This is it

Since $f$ is a continuous decreasing function,

$\begin{aligned} f(8) \le & f(6) \le f(4) \\ \frac {f(8)}{f(8)} \le & \frac {f(6)}{f(8)} \le \frac {f(4)}{f(8)} \\ \implies \lim_{x \to \infty} \frac {f(8)}{f(8)} \le \lim_{x \to \infty} & \frac {f(6)}{f(8)} \le \lim_{x \to \infty} \frac {f(4)}{f(8)} \\ 1 \le \lim_{x \to \infty} & \frac {f(6)}{f(8)} \le 1 \end{aligned}$

By squeeze or sandwich theorem , we have: $\displaystyle \implies \lim_{x \to \infty} \frac {f(6)}{f(8)} = \boxed{1}$

Visualise $f$ as $f(x)=1/log(x)$ as it satisfies $D_f$ , $R_f$ , the decreasing nature and

$\lim\limits_{x \to \infty} \frac {log4x} {log8x}$ = $\lim\limits_{x \to \infty} \frac {log4+logx} {log8+logx}$ = $\lim\limits_{x \to \infty} \frac {log4/logx+1}{log8/logx+1}$ = $1$

Similarly we can show that,

$\lim\limits_{x \to \infty} \frac {log6x} {log8x}$ = $1$ .

Its the sandwich theorem..