limt→011sin2t+21sin2t+31sin2t......n1sin2t^sin2t is equal to

Correct answer: (B) n. L=limt→011sin2t+21sin2t+31sin2t......n1sin2t^sin2t =limt→01cosec2t+2cosec2t+3cosec2t......ncosec2t^sin2t =limt→0 n^sin2tsin2t1n^cosec2t+2n^cosec2t+3n^cosec2t......1sin^2t =nlimt→0 0+0+0......1^0=n

Q58 / 90 · JEE Main 2023 (24 Jan Shift 1) · Mathematics

$\lim_{t \to 0} {(1^{\frac{1}{\sin^{2} t}} + 2^{\frac{1}{\sin^{2} t}} + 3^{\frac{1}{\sin^{2} t}} . . . . . . n^{\frac{1}{\sin^{2} t}})}^{\sin^{2} t}$ is equal to

A
$n^{2} + n$
Correct answer
B
$n$
C
$\frac{n (n + 1)}{2}$
D
$n^{2}$

Solution

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

$\underset{t \to 0}{= \lim} {(1^{{cosec}^{2} t} + 2^{{cosec}^{2} t} + 3^{{cosec}^{2} t} . . . . . . n^{{cosec}^{2} t})}^{\sin^{2} t}$

$\underset{t \to 0}{= \lim} n^{\frac{\sin^{2} t}{\sin^{2} t}} {({(\frac{1}{n})}^{{cosec}^{2} t} + {(\frac{2}{n})}^{{cosec}^{2} t} + {(\frac{3}{n})}^{{cosec}^{2} t} . . . . . . 1)}^{\sin^{2} t}$

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

Hence, the correct answer is option: 'B'.

limt→011sin2t+21sin2t+31sin2t......n1sin2tsin2t is equal to

Solution

Relevant Questions

$\lim_{t \to 0} {(1^{\frac{1}{\sin^{2} t}} + 2^{\frac{1}{\sin^{2} t}} + 3^{\frac{1}{\sin^{2} t}} . . . . . . n^{\frac{1}{\sin^{2} t}})}^{\sin^{2} t}$ is equal to