Question

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

• A. 4
• B. 0
• C. \(\frac{7}{5}\)
• D. 1

Answer 1

The Correct Option is B

Given:

\[ \lim_{x \to \infty} \frac{f(7x)}{f(x)} = 1 \]

Since \( f \) is strictly increasing, we have:

\[ f(x) < f(5x) < f(7x) \]

This implies:

\[ \lim_{x \to \infty} \frac{f(5x)}{f(x)} = 1 \]

Then:

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

Thus, the answer is: 0.