$$\lim_{t \to 0} \left(1^{\frac{1}{\sin^2 t}} + 2^{\frac{1}{\sin^2 t}} + 3^{\frac{1}{\sin^2 t}} \cdots n^{\frac{1}{\sin^2 t}}\right)^{\sin^2 t}$$ is equal to

We need to evaluate $$\lim_{t \to 0} \left(1^{1/\sin^2 t} + 2^{1/\sin^2 t} + 3^{1/\sin^2 t} + \cdots + n^{1/\sin^2 t}\right)^{\sin^2 t}$$.

As $$t \to 0$$, one has $$\sin^2 t \to 0$$ and hence $$\frac{1}{\sin^2 t} \to \infty$$. By setting $$s = \frac{1}{\sin^2 t}$$, the problem reduces to evaluating the limit $$\lim_{s \to \infty} \left(1^s + 2^s + 3^s + \cdots + n^s\right)^{1/s}$$.

For large $$s$$, the term $$n^s$$ dominates the sum since it grows faster than any other $$k^s$$ with $$k < n$$. Factoring out $$n^s$$ from inside the parentheses yields

$$\left(n^s\left[\left(\frac{1}{n}\right)^s + \left(\frac{2}{n}\right)^s + \cdots + 1\right]\right)^{1/s} = n \cdot \left[\left(\frac{1}{n}\right)^s + \cdots + 1\right]^{1/s}$$.

As $$s \to \infty$$, each term $$\left(\frac{k}{n}\right)^s \to 0$$ for $$k < n$$, so the bracketed sum approaches $$1$$, and the limit simplifies to $$n \cdot 1 = n$$.

The answer is Option 2: $$n$$.