Question

$\underset{t\to 0}{\lim }{\left(1^{\frac{1}{{\sin }^{2}t}}+2^{\frac{1}{{\sin }^{2}t}}+\dots +n^{\frac{1}{{\sin }^{2}t}}\right)}^{{\sin }^{2}t}$ is equal to

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

Answer 1

Answer: (C)

$\underset{t\to 0}{\lim }{\left(1^{{\text{cosec}}^{2}t}+2^{{\text{cosec}}^{2}t}+\dots \dots ..+n^{{\text{cosec}}^{2}t}\right)}^{{\sin }^{2}t}$
$=\underset{t\to 0}{\lim }n{\left({\left(\frac{1}{n}\right)}^{{\text{cosec}}^{2}t}+{\left(\frac{2}{n}\right)}^{{\text{cosecs}}^{2}t}+\dots \dots ..+1\right)}^{{\sin }^{2}t}$
$=n$