JEE Main 2021 · MathsMultiple choiceSingle correctHardCalculation

JEE Main 3 August 2021, Shift 1, Maths Q65

Question 65 of 90 in this shift, Maths question 5 of 30, Section A.

The value of the limit :

\lim_{x \to 4} \left(2^x + 2^{5-x} - 17\right)^{\frac{1}{8 - 2^{3x/4}}}

is equal to

(1) $e^{7/3}$
(2) $e^{-7/3}$ Official answer
(3) $e^{3/7}$
(4) $3e^7$

Official answer

Option 2

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

3 Apr 2025, Shift 2 · Q25If $\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = p$ , then $96 \log_e p$ is equal to ___________.MediumNumerical value
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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.