_ t → 0 (1 1 ^ 2 t + 2 1 ^ 2 t + 3 1 ^ 2 t . . . . . . n 1 ^ 2 t ) ^ ^ 2 t is equal to

Correct answer: B. n L = _ t → 0 (1 1 ^ 2 t + 2 1 ^ 2 t + 3 1 ^ 2 t . . . . . . n 1 ^ 2 t ) ^ ^ 2 t _ t → 0 ( 1 ^ cosec ^ 2 t + 2 ^ cosec ^ 2 t + 3 ^ cosec ^ 2 t . . . . . . n ^ cosec ^ 2 t ) ^ ^ 2 t _ t → 0 n ^ 2 t ^ 2 t ( ((1)/(n)) ^ cosec ^ 2 t + ((2)/(n)) ^ cosec ^ 2 t + ((3)/(n)) ^ cosec ^ 2 t . . . . . . 1) ^ ^ 2 t = n _ t → 0 (0 + 0 + 0 . . . . . . 1) ^ 0 = n

JEE Main 202324 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2023 · 24 January · Morning Shift · Mathematics · Limits

_ t → 0 (1 1 ^ 2 t + 2 1 ^ 2 t + 3 1 ^ 2 t . . . . . . n 1 ^ 2 t )...

\lim_{t \to 0}{

\left(1\dfrac{1}{\sin^{2}t} + 2\dfrac{1}{\sin^{2}t} + 3\dfrac{1}{\sin^{2}t} . . . . . . n\dfrac{1}{\sin^{2}t}\right)}^{\sin^{2}t}

is equal to

Options

A ${n}^{2} + n$
B $n$
C\dfrac{n $\left(n + 1\right)}{2}$
D ${n}^{2}$

Correct answer

B. $n$

Step-by-step solution

$L = \lim_{t \to 0}{\left(1\dfrac{1}{\sin^{2}t} + 2\dfrac{1}{\sin^{2}t} + 3\dfrac{1}{\sin^{2}t} . . . . . . n\dfrac{1}{\sin^{2}t}\right)}^{\sin^{2}t}$

$\lim_{t \to 0}{\left({1}^{{cosec}^{2}t} + {2}^{{cosec}^{2}t} + {3}^{{cosec}^{2}t} . . . . . . {n}^{{cosec}^{2}t}\right)}^{\sin^{2}t}$

$\lim_{t \to 0}n\dfrac{\sin^{2}t}{\sin^{2}t}{\left({\left(\dfrac{1}{n}\right)}^{{cosec}^{2}t} + {\left(\dfrac{2}{n}\right)}^{{cosec}^{2}t} + {\left(\dfrac{3}{n}\right)}^{{cosec}^{2}t} . . . . . . 1\right)}^{\sin^{2}t}$

$= n\lim_{t \to 0}{\left(0 + 0 + 0 . . . . . . 1\right)}^{0} = n$

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