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?

Let S be the set of all (α, β) ∈ R × R such that lim_(x → ∞) (sin(x²)(logₑ x)^α sin(1/x²))/(x^(αβ)(logₑ(1+x))^β) = 0 Then which of the following is (are) correct?

$S \subseteq \mathbb{R}^2:\ \lim_{x\to\infty} \dfrac{\sin(x^2)\,(\ln x)^\alpha\,\sin(1/x^2)}{x^{\alpha\beta}\,(\ln(1+x))^\beta} = 0$

$\text{four given pairs } (\alpha,\beta)$

$\text{Check: } (-1,3),\ (-1,1),\ (1,-1),\ (1,-2)\ \in S\ ?$

$\sin(1/x^2) \sim \dfrac{1}{x^2},\qquad \ln(1+x) \sim \ln x$

$|\sin(x^2)| \le 1 \quad (\text{bounded, oscillating})$

$\dfrac{(\ln x)^\alpha \cdot \frac{1}{x^2}}{x^{\alpha\beta}(\ln x)^\beta} = \dfrac{(\ln x)^{\alpha-\beta}}{x^{2+\alpha\beta}}$

$\text{expression} \sim \sin(x^2)\cdot\dfrac{(\ln x)^{\alpha-\beta}}{x^{2+\alpha\beta}}$

$\text{Since } \sin(x^2) \text{ oscillates: limit } = 0 \iff A(x) \to 0$

$A(x) = \dfrac{(\ln x)^{\alpha-\beta}}{x^{2+\alpha\beta}}$

$2 + \alpha\beta > 0 \Rightarrow A(x) \to 0$

$2 + \alpha\beta < 0 \Rightarrow A(x) \to \infty$

$2 + \alpha\beta = 0 \Rightarrow A(x) = (\ln x)^{\alpha-\beta} \to 0 \iff \alpha < \beta$

$S = \{(\alpha,\beta): \alpha\beta > -2\} \cup \{(\alpha,\beta): \alpha\beta = -2,\ \alpha < \beta\}$

$(1)\ (-1,3):\ \alpha\beta = -3 < -2 \ \times$

$(2)\ (-1,1):\ \alpha\beta = -1 > -2 \ \checkmark$

$(3)\ (1,-1):\ \alpha\beta = -1 > -2 \ \checkmark$

$(4)\ (1,-2):\ \alpha\beta = -2,\ \alpha > \beta \ \times$

$\boxed{\text{Correct options: } (2)\ \text{and}\ (3)}$