Let $f: \rightarrow \mathbb{R} \rightarrow(0, \infty)$ be strictly increasing function such that $\lim _\limits{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$. Then, the value of $\lim _\limits{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$ is equal to

<p>$$\begin{aligned} & f: R \rightarrow(0, \infty) \\ & \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1 \end{aligned}$$</p> <p>$\because \mathrm{f}$ is increasing</p> <p>$$\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x}) \\ & \because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(7 \mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 1<\lim _{\mathrm{x} \rightarrow \infty} \frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<1 \\ & \therefore\left[\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}-1\right] \\ & \Rightarrow 1-1=0 \end{aligned}$$</p>