Let, $x$ be any poistive real number.

$\Rightarrow x<5x<7x$

$\Rightarrow f\left(x\right)<f\left(5x\right)<f\left(7x\right)$

$\Rightarrow \frac{f\left(x\right)}{f\left(x\right)}<\frac{f\left(5x\right)}{f\left(x\right)}<\frac{f\left(7x\right)}{f\left(x\right)}$

$\Rightarrow 1<\frac{f\left(5x\right)}{f\left(x\right)}<\frac{f\left(7x\right)}{f\left(x\right)}$

$\Rightarrow \underset{x\to \infty }{\lim }1<\underset{x\to \infty }{\lim }\frac{f\left(5x\right)}{f\left(x\right)}<\underset{x\to \infty }{\lim }\frac{f\left(7x\right)}{f\left(x\right)}$

It is given that, $\underset{x\to \infty }{\lim }\frac{f\left(7x\right)}{f\left(x\right)}=1$

So, by using Sandwich theorem,

$\Rightarrow \underset{x\to \infty }{\lim }\frac{f\left(5x\right)}{f\left(x\right)}=1$

$\Rightarrow \underset{x\to \infty }{\lim }\left\{\frac{f\left(5x\right)}{f\left(x\right)}-1\right\}=0$