Let S be the set of all (alpha, beta) in math | vedclass

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(B, C) Given the limit: $\lim _{x \rightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$. Since $

Vedclass · Accepted Answer

$(1, -1) \in S$

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(B, C) Given the limit: $\lim _{x ightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$. Since $\sin(x^2)$ is bounded and $\sin(1/x^2) \approx 1/x^2$ as $x ightarrow \infty$,and $\log_e(1+x) \approx \log_e x$ as $x ightarrow \infty$,the expression behaves as: $\lim _{x ightarrow \infty} \frac{(\log_e x)^\alpha \cdot (1/x^2)}{x^{\alpha \beta} \cdot (\log_e x)^\beta} = \lim _{x ightarrow \infty} \frac{(\log_e x)^{\alpha-\beta}}{x^{\alpha \beta+2}} = 0$. For this limit to be $0$,the exponent of $x$ in the denominator must be positive,i.e.,$\alpha \beta + 2 > 0$,which implies $\alpha \beta > -2$. Checking the options: $(A) (-1, 3) \implies (-1)(3) = -3 
gtr -2$ (Incorrect). $(B) (-1, 1) \implies (-1)(1) = -1 > -2$ (Correct). $(C) (1, -1) \implies (1)(-1) = -1 > -2$ (Correct). $(D) (1, -2) \implies (1)(-2) = -2 
gtr -2$ (Incorrect). Thus,$(B)$ and $(C)$ are correct.

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $\lim _{x \rightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$. Then which of the following is (are) correct?

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