JEE Advanced 2024 Paper 2Question 1
Let $S$ be the set of all $(\alpha,\beta)\in\mathbb{R}\times\mathbb{R}$ such that $$\lim_{x\to\infty}\frac{\sin(x^2)(\log_e x)^{\alpha}\sin\left(\frac{1}{x^2}\right)}{x^{\alpha\beta}(\log_e(1+x))^{\beta}}=0.$$ Then which of the following is (are) correct?
Limits & Continuity
Solution
Concept. $\sin(x^2)$ is bounded but does not tend to $0$, so the limit is $0$ exactly when the rest tends to $0$. Formulas used. (1) $\sin\frac1{x^2}\sim\frac1{x^2}$, $\ln(1+x)\sim\ln x$ as $x\to\infty$(2) $\frac{(\ln x)^p}{x^q}\to0$ for $q>0$ (any $p$); $(\ln x)^p\to\infty$ for $p>0$Step 1 - Simplify. By (1), the expression behaves like $$\sin(x^2)\cdot(\ln x)^{\alpha-\beta}\cdot x^{-2-\alpha\beta}$$ It $\to0$ iff $(\ln x)^{\alpha-\beta}x^{-2-\alpha\beta}\to0$. Step 2 - (A) $(-1,3)$. $\alpha-\beta=-4$, $-2-\alpha\beta=1$: $\frac{x}{(\ln x)^4}\to\infty$. Not in $S$. Step 3 - (B) $(-1,1)$. $\alpha-\beta=-2$, power $-1$: $\frac1{x(\ln x)^2}\to0$. In $S$. Step 4 - (C) $(1,-1)$. $\alpha-\beta=2$, power $-1$: $\frac{(\ln x)^2}{x}\to0$ by (2). In $S$. Step 5 - (D) $(1,-2)$. $\alpha-\beta=3$, power $0$: $(\ln x)^3\to\infty$. Not in $S$. Answer: B,C