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
- 4
- 0
- \(\frac{7}{5}\)
- 1
The Correct Option is B
Approach Solution - 1
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
Approach Solution -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.
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