Question

Let \( S \) be the set of all \( (\alpha, \beta) \in \mathbb{R} \times \mathbb{R} \) such that \[ \lim_{x \to \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin\left(\frac{1}{x^2}\right)}{x^{\alpha\beta} (\log_e(1+x))^\beta} = 0. \] Then which of the following is (are) correct?

• A. \( (-1, 3) \in S \)
• B. \( (-1, 1) \in S \)
• C. \( (1, -1) \in S \)
• D. \( (1, -2) \in S \)

Hint:

Always simplify limits by approximating logarithmic and trigonometric terms for large values of \( x \).

Answer 1

The Correct Option is B

The given limit simplifies as: \[ \lim_{x \to \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin\left(\frac{1}{x^2}\right)}{x^{\alpha\beta} (\log_e(1+x))^\beta}. \] For \( x \to \infty \), expanding: \[ \sin\left(\frac{1}{x^2}\right) \approx \frac{1}{x^2}. \] So: \[ \lim_{x \to \infty} \frac{(\log_e x)^\alpha}{x^{\alpha\beta} (\log_e(1+x))^\beta} \cdot \frac{1}{x^2}. \] Let \( \log_e x = t \), then: \[ \frac{(\log_e x)^\alpha}{x^{\alpha\beta}} \sim t^{\alpha} e^{-t\alpha\beta}. \] For the limit to be \( 0 \), we require: \[ \alpha\beta + 2>0 \quad \text{or} \quad \alpha\beta>-2. \] Substituting the options: For \( (-1, 3): \alpha\beta = -3 \) (not valid). For \( (-1, 1): \alpha\beta = -1 \) (valid). For \( (1, -1): \alpha\beta = -1 \) (valid). For \( (1, -2): \alpha\beta = -2 \) (not valid).