Let S be the set of all $(\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->})\in <!--\; \in \; -->\mathbb{R}\times <!--\; \times \; -->\mathbb{R}$ such that

$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\sin <!--\; \; -->\left(x^2\right){({\log }_e<!--\; \; -->x)}^{\mathrm{\alpha <!--\; \alpha \; -->}}\sin <!--\; \; -->\left(\frac{1}{x^2}\right)}{x^{\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{\beta <!--\; \beta \; -->}}{({\log }_e<!--\; \; -->(1+x))}^{\mathrm{\beta <!--\; \beta \; -->}}}=0$

Then which of the following is (are) correct?

The correct answer is:
$(-1,1)\in S$

To evaluate the limit and determine the conditions on $\alpha$ and $\beta$, we analyze the behavior of the terms in the limit expression as $x\to \mathrm{\infty }$.

Let the expression in the limit be $L$:
$L=\frac{\sin \left(x^2\right){\left({\log }_{e}x\right)}^{\alpha }\sin \left(\frac{1}{x^2}\right)}{x^{\alpha \beta }{({\log }_e(1+x))}^{\beta }}$

We use the following asymptotic behaviors as $x\to \mathrm{\infty }$:
1. Since $\frac{1}{x^2}\to 0$, we have:
$\sin \left(\frac{1}{x^2}\right)\approx \frac{1}{x^2}$
2. For the logarithmic term in the denominator:
${\log }_e(1+x)={\log }_e\left[x(1+\frac{1}{x})\right]={\log }_{e}x+{\log }_e(1+\frac{1}{x})\approx {\log }_{e}x$

Substituting these into the limit expression gives the simplified form:
$L\approx \frac{\sin \left(x^2\right){\left(\log x\right)}^{\alpha }\cdot \frac{1}{x^2}}{x^{\alpha \beta }{\left(\log x\right)}^{\beta }}=\sin \left(x^2\right)\cdot x^{-(2+\alpha \beta )}{\left(\log x\right)}^{\alpha -\beta }$

Since $\sin (x^2)$ is bounded, the limit as $x\to \mathrm{\infty }$ will be $0$ if the power of $x$ dominates and drives the expression to zero. Specifically, if $2+\alpha \beta >0$, then $x^{-(2+\alpha \beta )}\to 0$, and because powers of $x$ dominate over logarithmic terms, the limit is $0$.

Let us test the option $(\alpha ,\beta )=(-1,1)$:
$2+\alpha \beta =2+(-1)(1)=1>0$

Evaluating the limit for these values:
$\underset{x\to \mathrm{\infty }}{lim}\frac{\sin (x^2)}{x{\left(\log x\right)}^2}=0$
This confirms that $(-1,1)\in S$.