The value of : lim x→ 0 [1^{1/sin^2 x} + 2^{1/sin^2 x} + …. + n^{1/sin^2 x}]^{sin^2 x}

Q: The value of $\large lim_{x \rightarrow 0} [1^{1/sin^2 x} + 2^{1/sin^2 x} + …. + n^{1/sin^2 x}]^{sin^2 x} $

(A) ∞

(B) 0

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

(D) n

Solution : Put $\frac{1}{sin^2 x} = t \ge 1 $

$\large lim_{t \rightarrow \infty} [1^t + 2^t + …. + n^t]^{1/t} $

$\large lim_{t \rightarrow \infty} n[(\frac{1}{n})^t + (\frac{2}{n})^t + …. + 1]^{1/t} $

= n[0 + 0 + … + 1]0 = n

Hence (D) is the correct answer.