Mathematics · 2025
JEE Main · 23 January 2025, Shift 2 · Q14
lim_x → ∞ ((2 x^2-3 x+5)(3 x-1)^(x/2))/((3 x^2+5 x+4) √((3 x+2)^x)) is equal to:
$\displaystyle \lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{2}{3 \sqrt{\mathrm{e}}}$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.