JEE Advanced 2023 Paper 1Question 1
Let $f:(0,1)\to\mathbb{R}$ be the function defined as $f(x)=\sqrt{n}$ if $x\in\left[\frac{1}{n+1},\frac1n\right)$ where $n\in\mathbb{N}$. Let $g:(0,1)\to\mathbb{R}$ be a function such that $\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}\,dt<g(x)<2\sqrt{x}$ for all $x\in(0,1)$. Then $\lim_{x\to0}f(x)g(x)$
Limits & Continuity
Solution
For $x\in[\frac1{n+1},\frac1n)$, $\frac1x-1<n\le\frac1x$, so $f(x)\sim\frac1{\sqrt x}$. The integral $\int_{x^2}^x\sqrt{\frac{1-t}{t}}dt\sim2\sqrt x$, so $g(x)\sim2\sqrt x$ by squeezing and $f(x)g(x)\to2$.