JEE 2023 — Mathematics PYQ

\begin{aligned}
 & f(x)\leq\mathrm{g}(x)\leq h(x) \\
 & \lim_{x\to a}f(x)\leq\lim_{x\to a}g(x)\leq\lim_{x\to a}h(x) \\
 & \mathrm{L}=\operatorname*{lim}_{x\to a}g(x)\leq\mathrm{L} \\
 & \mathrm{Let}2^{\frac{1}{2}}-2^{\frac{1}{3}}\to\mathrm{~smallest} \\
 & 2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\to\mathrm{~largest} \\
 & \left.\left(
\begin{array}
{cc}{\frac{1}{2}} & {\frac{1}{3}} \\
{2^{\frac{1}{2}}-2^{\frac{1}{3}}} \\
{}
\end{array}\right.\right)\leq p\leq
\begin{pmatrix}
{2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}} \\
{}
\end{pmatrix}^{n} \\
 & \lim_{n\to\infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^{n}\leq p\leq\lim_{n\to\infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)^{n} \\
 & 0=p
\end{aligned}