Question

$\displaystyle\lim _{t \rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots+n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ is equal to

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

Answer 1

The Correct Option is C

Rewrite the expression inside the limit: \[ \lim_{t \to 0} \left( \frac{1}{\sin^2 t} + \frac{1}{2 \sin^2 t} + \ldots + \frac{1}{n \sin^2 t} \right)^{\sin^2 t}. \] Factor out \( \frac{1}{\sin^2 t} \): \[ = \lim_{t \to 0} \left( \frac{1}{\sin^2 t} \left( 1 + \frac{1}{2} + \ldots + \frac{1}{n} \right) \right)^{\sin^2 t}. \] Simplify \( 1 + \frac{1}{2} + \ldots + \frac{1}{n} \) using the sum of harmonic series approximation: \[ 1 + \frac{1}{2} + \ldots + \frac{1}{n} \approx \ln n \, \text{(for large \( n \))}. \] Substitute: \[ \lim_{t \to 0} \left( \frac{1}{\sin^2 t} \cdot \ln n \right)^{\sin^2 t}. \] Using the property \(\lim_{t \to 0} (a^b) = e^{b \ln a}\): \[ \lim_{t \to 0} \left( \frac{\ln n}{\sin^2 t} \right)^{\sin^2 t} \to e^{\ln n} = n. \] Final Answer: \( n \).