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Limits Continuity And Differentiability

JEE Main 2023 (Online) 24th January Morning Shift

MCQ+4 / -12023

\(\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}\) is equal to

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