Set of Ordered Pairs for Which Specified Limit Equals Zero

To determine which ordered pairs $(\alpha, \beta)$ belong to the set $S$, we need to evaluate the limit: $L = \lim_{x \to \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin\left(\frac{1}{x^2}\right)}{x^{\alpha\beta} (\log_e (1+x))^\beta}$

Step 1: Asymptotic Behavior Analysis

As $x \to \infty$, we use the following standard asymptotic approximations:

1. $\sin\left(\frac{1}{x^2}\right) = \frac{1}{x^2}\left(1 + O\left(\frac{1}{x^4}\right)\right)$
2. $\log_e(1+x) = \log_e\left(x\left(1 + \frac{1}{x}\right)\right) = \log_e x + \log_e\left(1 + \frac{1}{x}\right) = \log_e x \left(1 + \frac{1}{x \log_e x} + o(1)\right)$

Substituting these into the expression inside the limit gives: $f(x) = \frac{\sin(x^2)(\log_e x)^\alpha \cdot \frac{1}{x^2}(1 + o(1))}{x^{\alpha\beta} (\log_e x)^\beta (1 + o(1))}$

Simplifying the algebraic expression: $f(x) = \sin(x^2) \cdot x^{-(\alpha\beta + 2)} (\log_e x)^{\alpha - \beta} (1 + o(1))$

Let $h(x) = x^{-(\alpha\beta + 2)} (\log_e x)^{\alpha - \beta}$. Then: $f(x) = \sin(x^2) \cdot h(x) \cdot (1 + o(1))$

Step 2: Condition for the Limit to equal Zero

Since $\sin(x^2)$ oscillates indefinitely between $-1$ and $1$ as $x \to \infty$, the overall limit $\lim_{x \to \infty} f(x) = 0$ holds if and only if: $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \frac{(\log_e x)^{\alpha - \beta}}{x^{\alpha\beta + 2}} = 0$

We analyze the behavior of $h(x)$ based on the exponent of $x$:

1. If $\alpha\beta + 2 > 0$: The power of $x$ in the denominator is strictly positive. Since polynomial growth dominates logarithmic growth ($\lim_{x \to \infty} \frac{(\log_e x)^k}{x^p} = 0$ for any $p > 0$ and $k \in \mathbb{R}$), we have $\lim_{x \to \infty} h(x) = 0$. Thus, $(\alpha, \beta) \in S$.
2. If $\alpha\beta + 2 = 0$: We have $h(x) = (\log_e x)^{\alpha - \beta}$, which approaches $0$ if and only if $\alpha - \beta < 0$.
3. If $\alpha\beta + 2 < 0$: The term $x^{-(\alpha\beta + 2)}$ grows polynomially to $\infty$, dominating any logarithm in the denominator, so $h(x) \to \infty \neq 0$. Thus, $(\alpha, \beta) \notin S$.

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Step 3: Checking the Given Options

• Option (A): $(-1, 3)$ Here, $\alpha = -1$ and $\beta = 3$. $\alpha\beta + 2 = (-1)(3) + 2 = -1 < 0$ Therefore, $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \frac{x}{(\log_e x)^4} = \infty \neq 0$. So, $(-1, 3) \notin S$. (Option A is incorrect)

• Option (B): $(-1, 1)$ Here, $\alpha = -1$ and $\beta = 1$. $\alpha\beta + 2 = (-1)(1) + 2 = 1 > 0$ Therefore, $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \frac{1}{x(\log_e x)^2} = 0$. So, $(-1, 1) \in S$. (Option B is correct)

• Option (C): $(1, -1)$ Here, $\alpha = 1$ and $\beta = -1$. $\alpha\beta + 2 = (1)(-1) + 2 = 1 > 0$ Therefore, $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \frac{(\log_e x)^2}{x} = 0$. So, $(1, -1) \in S$. (Option C is correct)

• Option (D): $(1, -2)$ Here, $\alpha = 1$ and $\beta = -2$. $\alpha\beta + 2 = (1)(-2) + 2 = 0$ Checking $\alpha - \beta$: $1 - (-2) = 3 > 0$. Therefore, $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} (\log_e x)^3 = \infty \neq 0$. So, $(1, -2) \notin S$. (Option D is incorrect)

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Final Conclusion

The correct options are (B) and (C).