Mathematics · 2025
JEE Main · 7 April 2025, Shift 1 · Q16
lim_x → 0^+ (tan (5(x)^(1/3)) log_e(1+3 x^2))/(( tan^-1 3 √ x )^2(e^5(x)^(4/3) -1)) is equal to
$\displaystyle \lim _{x \rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\frac{4}{3}}}-1\right)}$ is equal to
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{1}{3}$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.