Let \( f:\mathbb{R}\to (0,\infty) \) be strictly increasing function such that \( \displaystyle\lim_{x\to\infty}\dfrac{f(7x)}{f(x)}=1 \). Then, the value of \( \displaystyle \lim_{x\to\infty}\left[\dfrac{f(5x)}{f(x)}-1\right] \) is equal to:
\( 0 \)
\( 1 \)
\( \dfrac{7}{5} \)
\( 4 \)
