Mathematics · 2023
JEE Main · 24 January 2023, Shift 1 · Q74
lim_t → 0(1^(1/(sin^2 t))+2^(1/(sin^2 t))+…+n^(1/(sin^2 t)))^sin^2 t is equal to
$\displaystyle \lim _{t \rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots+n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ is equal to
Official answer
From NTA’s final answer key for this paper.
(4)
n
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.