Question 5

To analyze the limit $L = \lim_{x \to \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha\beta} (\log_e(1+x))^\beta}$, we consider the behavior of each term as $x \to \infty$:

1. $\sin(x^2)$ is a bounded function in the interval $[-1, 1]$.
2. $\sin\left(\frac{1}{x^2}\right) \approx \frac{1}{x^2}$ for large $x$.
3. $\log_e(1+x) = \log_e(x(1 + 1/x)) = \log_e x + \log_e(1 + 1/x) \approx \log_e x$ for large $x$.

Substituting these approximations into the expression: $L \approx \lim_{x \to \infty} \frac{\sin(x^2) (\log_e x)^\alpha \cdot \frac{1}{x^2}}{x^{\alpha\beta} (\log_e x)^\beta} = \lim_{x \to \infty} \frac{\sin(x^2) (\log_e x)^{\alpha - \beta}}{x^{\alpha\beta + 2}}$ For the limit to be $0$, the power of $x$ in the denominator must dominate. This happens if: i) $\alpha\beta + 2 > 0$ (since algebraic growth $x^k$ dominates logarithmic growth $(\log x)^m$ for $k > 0$). ii) If $\alpha\beta + 2 = 0$, then we must have the power of the log term $\alpha - \beta < 0$ so that the expression vanishes.

Evaluating the options:

• (A) $(-1, 3): \alpha\beta + 2 = (-1)(3) + 2 = -1 < 0$. The limit tends to $\infty$. (Incorrect)
• (B) $(-1, 1): \alpha\beta + 2 = (-1)(1) + 2 = 1 > 0$. The limit is $0$. (Correct)
• (C) $(1, -1): \alpha\beta + 2 = (1)(-1) + 2 = 1 > 0$. The limit is $0$. (Correct)
• (D) $(1, -2): \alpha\beta + 2 = (1)(-2) + 2 = 0$. Here $\alpha - \beta = 1 - (-2) = 3$. The limit becomes $\lim_{x \to \infty} \sin(x^2) (\log_e x)^3$, which oscillates and diverges. (Incorrect)