Q38 / 75 · JEE Main 2025 (3 Apr Shift 2) · Mathematics
If $\operatorname{Lim}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}=p$, then $96 \log _e p$ is equal to ______
Correct answer: 32
Solution
$P=\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$
$\Rightarrow \mathrm{P}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 0}\left(\frac{\tan x-\mathrm{x}}{\mathrm{x}^3}\right)}$
$=\mathrm{e}^{\lim _{x \rightarrow 0} \frac{\left(x+\frac{x^3}{3}+\frac{2 x^5}{15}+\ldots . .-x\right)}{x^3}}$
$=e^{1 / 3}$
$\therefore 96 \log _{\mathrm{e}}^{\mathrm{p}}=96 \times \frac{1}{3}=32$
$\Rightarrow \mathrm{P}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 0}\left(\frac{\tan x-\mathrm{x}}{\mathrm{x}^3}\right)}$
$=\mathrm{e}^{\lim _{x \rightarrow 0} \frac{\left(x+\frac{x^3}{3}+\frac{2 x^5}{15}+\ldots . .-x\right)}{x^3}}$
$=e^{1 / 3}$
$\therefore 96 \log _{\mathrm{e}}^{\mathrm{p}}=96 \times \frac{1}{3}=32$
Hence, the correct answer is 32.