JEE Main 2025 — Limits Question with Solution
JEE Main 2025 (23 Jan Shift 2)
Question
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Correct answer
C
Step-by-step explanation
$\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{x / 2}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}} \\
& =\lim _{x \rightarrow \infty} \frac{x^2\left(2-\frac{3}{x}+\frac{5}{x^2}\right)(3 x)^{\frac{x}{2}}}{x^2\left(3+\frac{5}{x}+\frac{4}{x^2}\right)(3 x)^{\frac{x}{2}}} \cdot \frac{\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}}{\left(1+\frac{2}{3 x}\right)^{\frac{x}{2}}} \\
& \lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}\right)^{\frac{x}{2}}=e^{\lim _{x \rightarrow \infty}\left(1-\frac{1}{3 x}-1\right) \times \frac{x}{2}}=e^{\frac{-1}{6}} \\
& \lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}\right)=e^{\lim _{x \rightarrow \infty}\left(1+\frac{2}{3 x}-1\right) \times \frac{x}{2}} \\
& =e^{\frac{1}{3}}
\end{aligned}$
So,
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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.