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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
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
Updated On: Nov 19, 2024
- 4
- 0
- \(\frac{7}{5}\)
- 1
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The Correct Option is B
Solution and Explanation
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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