lim t 0 ( 1 1 sin 2 t + 2 1 sin 2 t + ? + n 1 sin 2 t ) sin 2 t is equal to [2023]

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Solution

Q.
$\lim_{t \to 0} {(1^{\frac{1}{^{\sin^{2} t}}} + 2^{\frac{1}{^{\sin^{2} t}}} + \dots + n^{\frac{1}{^{\sin^{2} t}}})}^{\sin^{2} t}$ is equal to [2023]

1 $n^{2}$

2 $n^{2} + n$

3 $n$

4 $\frac{n (n + 1)}{2}$

Ans.
(3)

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

$\Rightarrow L = \lim_{t \to 0} n {[{(\frac{1}{n})}^{c o s e c^{2} t} + {(\frac{2}{n})}^{c o s e c^{2} t} + \dots + {(\frac{n - 1}{n})}^{c o s e c^{2} t} + 1]}^{\sin^{2} t}$

$\Rightarrow L = n (1^{\sin (0)}) = n (1) = n [0 < \frac{k}{n} < 1, {(\frac{k}{n})}^{\infty} = 0 for 1 \leq k < n]$

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