JEE Main 2025 · MathsNumerical answerNumerical valueMediumCalculation

JEE Main 3 April 2025, Shift 2, Maths Q25

Question 25 of 75 in this shift, Maths question 25 of 25, Section B.

\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = p

, then

96 \log_e p

is equal to ___________.

Official answer

NTA final key.

Chapter: Limits and Derivatives
Topic: L'Hopital's Rule
Idea tested: Indeterminate Power Forms

Same idea in other shifts

Asked 5× in all

9 Apr 2024, Shift 2 · Q5 $\lim_{x\to0}\frac{e-(1+2x)^{\frac{1}{2x}}}{x}$ is equal toHardSingle correct
3 Aug 2021, Shift 1 · Q65The value of the limit : $\lim_{x \to 4} \left(2^x + 2^{5-x} - 17\right)^{\frac{1}{8 - 2^{3x/4}}}$ is equal toHardSingle correct
3 Aug 2021, Shift 2 · Q68 $\lim_{x\to0}(1+x)^{\frac{1}{e^x-1}}$ is equal toEasySingle correct
31 Aug 2021, Shift 2 · Q67If $\alpha = \lim_{x\to\pi/4}\frac{\tan^3 x-\tan x}{\cos\left(x+\frac{\pi}{4}\right)}$ and $\beta = \lim_{x\to 0}(\cos x)^{\cot x}$ are…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.