$\text{Let }\mathrm{S }\text{be the set of all}(\alpha ,\beta )\in R\times R\text{such that}$

$\underset{x\to \infty }{lim}\frac{\sin \left(x^2\right){\left({\log }_{e}x\right)}^{\alpha }\sin \left(\frac{1}{x^2}\right)}{x^{\beta }{\left({\log }_e\right(1+x\left)\right)}^{\beta }}=0.$

$\text{Then which of the following is(are) correct ?}$

The correct options are $(-1,1)\in$ (-1, 1) ∈ S and (1, -1) ∈ S.

Step 1: Simplify the asymptotic behavior of the functions as x → ∞
As $x\to \infty$ (i.e. $x\to \infty$):
1. $\frac{1}{x^2}\to 0$, so using the standard limit $\underset{t\to 0}{lim}\frac{\sin \left(t\right)}{t}=1$, we have:

$\sin \left(\frac{1}{x^2}\right)\sim \frac{1}{x^2}$

2. For the logarithmic term in the denominator:

${\log }_e(1+x)={\log }_e\left(x\right(1+\frac{1}{x}\left)\right)={\log }_{e}x+{\log }_e(1+\frac{1}{x})\sim {\log }_{e}x$

Step 2: Rewrite the overall expression
Substituting these asymptotic approximations into the given limit expression:

$\underset{x\to \infty }{lim}\frac{\sin \left(x^2\right){\left({\log }_{e}x\right)}^{\alpha }\cdot <!--\; dot\; -->\frac{1}{x^2}}{x^{\beta }{\left({\log }_{e}x\right)}^{\beta }}=\underset{x\to \infty }{lim}\sin \left(x^2\right)\cdot x^{-(2+\beta )}\cdot {\left({\log }_{e}x\right)}^{\alpha -\beta }$

Step 3: Analyze the convergence condition for the limit
Since $\sin \left(x^2\right)$ is a bounded function oscillating continuously in the range $[-1,1]$, the entire limit will evaluate to $0$ if and only if the remaining algebraic factor decays to zero:

$\underset{x\to \infty }{lim}{x}^{-(2+\beta )}{\left({\log }_{e}x\right)}^{\alpha -\beta }=0$

For polynomial-versus-logarithmic growth rates as $x\to \infty$:
- If $-(2+\beta )<0\Rightarrow \beta >-<!--;</mo--><mn>2</mn>$, the negative exponent of $x$ places it in the denominator with a positive power. Since power functions grow strictly faster than any logarithmic power, the limit is $0$ for any value of $\alpha \in R$.

Step 4: Check the given options
1. For option (-1, 1): Here $\alpha =-1$ and $\beta =1$. Since $\beta =1>-2$, it satisfies the condition. Hence, $(-1,1)\in <!--\; in\; -->S$.
2. For option (1, -1): Here $\alpha =1$ and $\beta =-1$. Since $\beta =-1>-2$, it also satisfies the condition. Hence, $(1,-1)\in <!--\; in\; -->S$.