Limits Question 1
Question 1 - 24 January - Shift 1
$\lim_{t \to 0}(1^{\frac{1}{\sin ^{2} t}}+2^{\frac{1}{\sin ^{2} t}}+\ldots .+n^\frac{1}{sin^2t})^{\frac{1}{\sin ^{2} t}}$ is equal to
(1) $n^{2}+n$
(2) $n$
(3) $\frac{n(n+1)}{2}$
(4) $n^{2}$
Show Answer
Answer: (2)
Solution:
Formula: L’Hospital’s rule
$\begin{aligned} & \lim _{t \rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots \ldots . .+n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t} \\ & =\lim _{t \rightarrow 0} n\left(\left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t}+\left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t}+\ldots \ldots .+1\right)^{\sin ^2 t} \\ & =n \end{aligned}$
