A calculus problem by Prajwal Krishna

Take n^(1/sin^2(x)) common

As x is approaching zero so power is tending towards infinity.

All numbers except 1 left inside after taking common are less than 1 hence they will tend to zero

Hence the limit is n

$\begin{aligned} L & = \lim_{x \to 0} \left(1^{1/\sin^2 x} + 2^{1/\sin^2 x} + 3^{1/\sin^2 x} +...+n^{1/\sin^2 x} \right)^{\sin^2 x} & \small {\color{#3D99F6}\text{Let }u=\frac 1{\sin^2 x}} \\ & = \lim_{u \to \infty} \left(1+2^u+3^u+...+n^u \right)^{\frac 1u} \\ & = \lim_{u \to \infty} \exp \left(\ln \left(1+2^u+3^u+...+n^u \right)^{\frac 1u} \right) \\ & = \exp \left(\lim_{u \to \infty} \frac {\ln \left(1+2^u+3^u+...+n^u \right)}u \right) \\ & = \exp \left(\lim_{u \to \infty} \frac {\ln \left(n^u \left(\frac {1+2^u+3^u+...+(n-1)^u}{n^u}+1\right) \right)}u \right) \\ & = \exp \left(\lim_{u \to \infty} \frac {\ln n^u + \ln \left(\frac {1+2^u+3^u+...+(n-1)^u}{n^u}+1\right)}u \right) \\ & = e^{\ln n} = \boxed{n} \end{aligned}$