$\mathrm{f}:(1,\mathrm{\infty })\to (0,\mathrm{\infty })$ be a continuous decreasing function $\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \frac{\mathrm{f}\left(4\mathrm{x}\right)}{\mathrm{f}\left(8\mathrm{x}\right)}=1$, Then $\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \frac{\mathrm{f}\left(6\mathrm{x}\right)}{\mathrm{f}\left(8\mathrm{x}\right)}\text{ is equal to }=$_______ 

Given that the function $f\left(x\right)$ is contiuous and decreasing function in the interval $\left(0,\infty \right)$

Hence, $f\left(8x\right)<f\left(6x\right)<f\left(4x\right)$

It implies that 

$\begin{array}{rcl}\underset{x\to 0}{\lim }1& <& \underset{x\to 0}{\lim }\frac{f\left(6x\right)}{f\left(8x\right)}<\underset{x\to 0}{\lim }\frac{f\left(4x\right)}{f\left(8x\right)}\\ \underset{x\to 0}{\lim }1& <& \underset{x\to 0}{\lim }\frac{f\left(6x\right)}{f\left(8x\right)}<1\end{array}$

Therefore $\underset{x\to 0}{\lim }\frac{f\left(6x\right)}{f\left(8x\right)}=1$