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

To solve this problem, we need to evaluate the limit:

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

The given condition is:

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

This tells us that for large values of \(x\), the function's growth rate satisfies:

\(f(7x) \approx f(x)\)

Since \(f(x)\) is strictly increasing and the condition holds for \(7x\), it indicates that \(f(x)\) grows in such a manner that scaling by any constant \(c > 0\) still results in the function growing similarly. Therefore, we assume:

\(\lim_{x \to \infty} \frac{f(cx)}{f(x)} = 1\) for any constant value of \(c\).

Now, applying this reasoning to the case when \(c = 5\), we have:

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

We can plug this into our original limit:

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

Given:

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

Thus, the expression simplifies to:

\(1 - 1 = 0\)

Therefore, the value of the limit is:

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.