1. Given

   We have a strictly increasing function f:(0,)\to \mathbb{R} with values in $(0,\infty)$, and $\lim_{x\to\infty}\frac{f(7x)}{f(x)}=1.$

   We need to find $\lim_{x\to\infty}\left[\frac{f(5x)}{f(x)}-1\right].$

2. Use monotonicity of $f$

   Since $f$ is strictly increasing and $5x<7x$ for all $x>0$, we have $f(x) < f(5x) < f(7x).$

   Because $f(x)>0$, we can divide throughout by $f(x)$: $1<\frac{f(5x)}{f(x)}<\frac{f(7x)}{f(x)}.$

3. Take limits

   We are given $\lim_{x\to\infty}\frac{f(7x)}{f(x)}=1.$

   So from $1<\frac{f(5x)}{f(x)}<\frac{f(7x)}{f(x)},$ the squeeze theorem gives $\lim_{x\to\infty}\frac{f(5x)}{f(x)}=1.$

4. Compute the required limit

   Therefore, $\lim_{x\to\infty}\left[\frac{f(5x)}{f(x)}-1\right]=1-1=0.$

5. Check options

   • A: $0$  correct
   • B: $4$  incorrect
   • C: $1$  incorrect
   • D: $\frac75$  incorrect

Hence the correct answer is A.