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(B) Given that $f: R \rightarrow (0, \infty)$ is a strictly increasing function.We are given the limit $\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x

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(B) Given that $f: R \rightarrow (0, \infty)$ is a strictly increasing function.We are given the limit $\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$.Since $f$ is a strictly increasing function and $x  0$,we have $f(x) Dividing the inequality by $f(x) > 0$,we get $1 Taking the limit as $x \rightarrow \infty$,we have $\lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(5x)}{f(x)} \leq \lim _{x \rightarrow \infty} \frac{f(7x)}{f(x)}$.Substituting the given limit,we get $1 \leq \lim _{x \rightarrow \infty} \frac{f(5x)}{f(x)} \leq 1$.By the Squeeze Theorem,$\lim _{x \rightarrow \infty} \frac{f(5x)}{f(x)} = 1$.Therefore,the value of $\lim _{x \rightarrow \infty} \left[\frac{f(5x)}{f(x)} - 1\right] = 1 - 1 = 0$.

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

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