Step-by-step Derivations:

1. Analyze the Limit Expression

   Let the given limit be $L$. The expression inside the limit is: $f(x) = \frac{\sin \left(x^2\right)\left(\log _e x\right)^\alpha \sin \left(\frac{1}{x^2}\right)}{x^{\alpha \beta}\left(\log _e(1+x)\right)^\beta}$ We need to find the set $S$ of all pairs $(\alpha, \beta)$ for which $\lim_{x \to \infty} f(x) = 0$.

2. Use Asymptotic Equivalents for $x \rightarrow \infty$

   For large values of $x$, we can use the following standard limits and approximations:

   • As $x \rightarrow \infty$, $\frac{1}{x^2} \rightarrow 0$. Therefore, $\sin\left(\frac{1}{x^2}\right) \sim \frac{1}{x^2}$. This is because $\lim_{u \to 0} \frac{\sin u}{u} = 1$.
   • As $x \rightarrow \infty$, $\log_e(1+x) = \log_e(x(1+1/x)) = \log_e x + \log_e(1+1/x)$. Since $\log_e(1+1/x) \to 0$, we have $\log_e(1+x) \sim \log_e x$. More formally, $\lim_{x \to \infty} \frac{\log_e(1+x)}{\log_e x} = 1$.

3. Simplify the Expression

   Based on these equivalents, the behavior of $f(x)$ for large $x$ is similar to the behavior of the following expression: $g(x) = \frac{\sin \left(x^2\right)\left(\log _e x\right)^\alpha \left(\frac{1}{x^2}\right)}{x^{\alpha \beta}\left(\log _e x\right)^\beta}$ Simplifying $g(x)$: $g(x) = \sin(x^2) \cdot \frac{(\log_e x)^{\alpha - \beta}}{x^{2+\alpha\beta}}$ The limit of $f(x)$ is 0 if and only if the limit of $g(x)$ is 0, because $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{\sin(1/x^2)}{1/x^2} \cdot \left(\frac{\log_e x}{\log_e(1+x)}\right)^\beta = 1 \cdot 1^\beta = 1$.

4. Determine the Conditions for the Limit to be Zero

   We need to find when $L = \lim_{x \rightarrow \infty} \sin(x^2) \cdot \frac{(\log_e x)^{\alpha-\beta}}{x^{2+\alpha \beta}} = 0$. The term $\sin(x^2)$ is a bounded function that oscillates between -1 and 1. For the overall limit to be zero, the magnitude of the other part must tend to zero. This is a direct application of the Squeeze Theorem. So, the condition is: $\lim_{x \rightarrow \infty} \frac{(\log_e x)^{\alpha-\beta}}{x^{2+\alpha \beta}} = 0$ We analyze this limit based on the power of $x$ in the denominator, which is $2+\alpha\beta$.

   • Case 1: $2 + \alpha\beta > 0$ In this case, the power of $x$ in the denominator is positive. Since polynomial functions grow faster than logarithmic functions, the limit will be 0, regardless of the value of the exponent $\alpha-\beta$. So, if $2 + \alpha\beta > 0$, then $(\alpha, \beta) \in S$.

   • Case 2: $2 + \alpha\beta < 0$ Let $2 + \alpha\beta = -k$ where $k > 0$. The expression becomes $\lim_{x \rightarrow \infty} x^k (\log_e x)^{\alpha-\beta}$. This limit is $\infty$, not 0. So these pairs are not in $S$.

   • Case 3: $2 + \alpha\beta = 0$ The expression becomes $\lim_{x \rightarrow \infty} (\log_e x)^{\alpha-\beta}$. For this limit to be 0, the exponent must be negative. So, we need $\alpha - \beta < 0$. If $\alpha - \beta \ge 0$, the limit is $\infty$ or 1, and the original limit $L$ would not be 0 (it would not exist).

   Summary of Conditions: A pair $(\alpha, \beta)$ is in the set $S$ if and only if: (i) $2 + \alpha\beta > 0$ OR (ii) $2 + \alpha\beta = 0$ AND $\alpha - \beta < 0$.

5. Evaluate Each Option

   • A: $(\alpha, \beta) = (-1, 3)$ $2 + \alpha\beta = 2 + (-1)(3) = 2 - 3 = -1$. Since $-1 < 0$, condition (i) is not met. Condition (ii) is also not met. Thus, $(-1,3) \notin S$.

   • B: $(\alpha, \beta) = (-1, 1)$ $2 + \alpha\beta = 2 + (-1)(1) = 2 - 1 = 1$. Since $1 > 0$, condition (i) is met. Thus, $(-1,1) \in S$.

   • C: $(\alpha, \beta) = (1, -1)$ $2 + \alpha\beta = 2 + (1)(-1) = 2 - 1 = 1$. Since $1 > 0$, condition (i) is met. Thus, $(1,-1) \in S$.

   • D: $(\alpha, \beta) = (1, -2)$ $2 + \alpha\beta = 2 + (1)(-2) = 2 - 2 = 0$. Condition (i) is not met. We check condition (ii). We have $2 + \alpha\beta = 0$. Now check the second part of condition (ii): $\alpha - \beta = 1 - (-2) = 3$. Since $3$ is not less than 0, the condition $\alpha - \beta < 0$ is not met. Thus, $(1,-2) \notin S$.

Conclusion:

The pairs $(\alpha, \beta)$ that belong to the set $S$ are $(-1,1)$ and $(1,-1)$. Therefore, options B and C are correct.