Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that
$$\lim_{x \to \infty} \dfrac{\sin(x^2)(\log_e x)^\alpha \sin\left(\dfrac{1}{x^2}\right)}{x^{\alpha\beta}(\log_e(1+x))^\beta} = 0.$$
Then which of the following is (are) correct?
Step-by-Step Solution
Key Concept: Asymptotic behavior of trigonometric and logarithmic functions at infinity, and the condition for a product of bounded and decaying functions to vanish.
As $x \to \infty$, we use the asymptotic approximations:
$$\sin\left(\dfrac{1}{x^2}\right) \sim \dfrac{1}{x^2} \quad \text{and} \quad \log_e(1+x) \sim \log_e x$$
The limit expression reduces to:
$$\lim_{x \to \infty} \sin(x^2) \dfrac{(\log_e x)^{\alpha - \beta}}{x^{2 + \alpha\beta}}$$
Since $\sin(x^2)$ is a bounded oscillating function, the limit is 0 if the power of $x$ in the denominator is strictly positive, i.e.,
$$2 + \alpha\beta > 0$$
Let's test the given options:
- For Option A: $(-1, 3) \implies 2 + (-1)(3) = -1 < 0$ (incorrect).
- For Option B: $(-1, 1) \implies 2 + (-1)(1) = 1 > 0$ (correct).
- For Option C: $(1, -1) \implies 2 + (1)(-1) = 1 > 0$ (correct).
- For Option D: $(1, -2) \implies 2 + (1)(-2) = 0$. Since $2+\alpha\beta=0$, the limit becomes $\lim_{x \to \infty} \sin(x^2) (\log_e x)^3 = \infty$ (incorrect).
Thus, the correct options are B and C.
Correct Answer: B, C