JEE Main 2025 — Limits Question with Solution
JEE Main 2025 (8 Apr Shift 2)
Question
Given below are two statements :
Statement I :
Statement II :
In the light of the above statements, choose the correct answer from the options given below :
Statement I :
Statement II :
In the light of the above statements, choose the correct answer from the options given below :
Choose an option
Show full solutionCorrect option: D
Correct answer
DBoth Statement I and Statement II are true
Step-by-step explanation
$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{3} \ldots-\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)\right]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots-2 x}{x^5}=\frac{2}{5} \\ & \lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow l}\left(\frac{2}{(1-x)}\right)(x-1)}=e^{-2}
\end{aligned}\Rightarrow$ Both statements correct
& \lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{3} \ldots-\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)\right]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots-2 x}{x^5}=\frac{2}{5} \\ & \lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow l}\left(\frac{2}{(1-x)}\right)(x-1)}=e^{-2}
\end{aligned}\Rightarrow$ Both statements correct
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This is a previous-year question from JEE Main 2025, covering the Limits chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.