JEE Main 2023 (Online) 6th April Evening ShiftQuestion 1
$\lim_{n\to\infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)\cdots\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}$ is equal to
Limits & Continuity
Solution
Each factor $2^{1/2}-2^{1/(2k+1)}$ is positive and for $k\ge1$ lies in $(0,\sqrt2-1]\subset(0,0.42)$ since $2^{1/(2k+1)}\ge1$. So the product of $n$ factors is $<(0.42)^{n-1}\cdot(\text{const})\to0$. Limit $=0$. Option (B).