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

Answer: (B)

0

Answer 2

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.