Mathematics · 2022
JEE Main · 28 July 2022, Shift 1 · Q29
lim_x → 0(((x+2 cos x)^3+2(x+2 cos x)^2+3 sin (x+2 cos x))/((x+2)^3+2(x+2)^2+3 sin (x+2)))^(100/x) is equal to \ \ \ \
$\displaystyle \lim _{x \rightarrow 0}\left(\frac{(x+2 \cos x)^3+2(x+2 \cos x)^2+3 \sin (x+2 \cos x)}{(x+2)^3+2(x+2)^2+3 \sin (x+2)}\right)^{\frac{100}{x}}$ is equal to
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Official answer
1
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.