JEE Main 2025 Apr 3 Shift 2, Mathematics Q23: Methods to Evaluate Limits

This is a $1^{\infty }$ indeterminate form. Use the standard limit ${\lim }_{x\to 0}(1+f(x){)}^{g(x)}=e^{\lim f(x)g(x)}$ with $f(x)=\frac{\tan x}{x}-1$ and $g(x)=\frac{1}{x^2}$.

Series expansion: $\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots$, so $\frac{\tan x}{x}-1=\frac{x^2}{3}+\frac{2x^4}{15}+\dots$.

$\underset{x\to 0}{\lim }\frac{1}{x^2}\left(\frac{\tan x}{x}-1\right)=\frac{1}{3}$.

Therefore $p=e^{1/3}$ and ${\log }_{e}p=1/3$.

$96{\log }_{e}p=96\cdot \frac{1}{3}=32$.