Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $\lim _{x \rightarrow…

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $\lim _{x \rightarrow \infty} \frac{\sin \left(x^2\right)\left(\log _e x\right)^\alpha \sin \left(\frac{1}{x^2}\right)}{x^{\alpha \beta}\left(\log _e(1+x)\right)^\beta}=0$. Then which of the following is (are) correct?
  1. $(-1,3) \in S$
  2. $(-1,1) \in S$
  3. $(1,-1) \in S$
  4. $(1,-2) \in S$

Solution

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\sin x^2 \cdot\left(\log _e x\right)^\alpha \cdot \sin \frac{1}{x^2}}{x^{\alpha \beta} \cdot\left(\log _e(1+x)\right)^\beta}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^\alpha}{\left(\log _e(x+1)\right)^\beta \cdot x^{\alpha \beta+2}}=0\end{aligned}$ $\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{\log _e x}{\log _e(x+1)}\right)^\beta \cdot \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \quad \text { Put } \log _e x=t \\ & \lim _{t \rightarrow \infty} \frac{t^{\alpha-\beta}}{\left(e^t\right)^{\alpha \beta+2}}=0\end{aligned}$ As we know $\lim _{x \rightarrow \infty} \frac{x}{e^x}=0$ $\alpha \beta+2>0 \Rightarrow \alpha \beta>-2$

Asked in: JEE Advanced 2024 (Paper 2)

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