$$\lim_{t \to 0} \left(1^{\frac{1}{\sin^2 t}} + 2^{\frac{1}{\sin^2 t}} + 3^{\frac{1}{\sin^2 t}} + \cdots + n^{\frac{1}{\sin^2 t}}\right)^{\sin^2 t}$$ 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 $$t \to 0$$, $$\sin t \to 0$$, so $$\frac{1}{\sin^2 t} \to \infty$$. Let us denote $$s = \frac{1}{\sin^2 t}$$, so $$s \to \infty$$ and $$\sin^2 t = \frac{1}{s}$$. The expression becomes:

$$L = \lim_{s \to \infty} \left(1^s + 2^s + 3^s + \cdots + n^s\right)^{1/s}$$

Among the terms $$1^s, 2^s, \ldots, n^s$$, the largest term is $$n^s$$ since $$n$$ is the largest base and $$s \to \infty$$. Factor out $$n^s$$:

$$L = \lim_{s \to \infty} \left[n^s\left(\left(\frac{1}{n}\right)^s + \left(\frac{2}{n}\right)^s + \cdots + \left(\frac{n-1}{n}\right)^s + 1\right)\right]^{1/s}$$ $$= \lim_{s \to \infty} n \cdot \left[\left(\frac{1}{n}\right)^s + \left(\frac{2}{n}\right)^s + \cdots + \left(\frac{n-1}{n}\right)^s + 1\right]^{1/s}$$

For each $$k < n$$, we have $$\frac{k}{n} < 1$$, so $$\left(\frac{k}{n}\right)^s \to 0$$ as $$s \to \infty$$. Therefore, the sum inside the brackets approaches:

$$\left(\frac{1}{n}\right)^s + \left(\frac{2}{n}\right)^s + \cdots + \left(\frac{n-1}{n}\right)^s + 1 \to 0 + 0 + \cdots + 0 + 1 = 1$$

More precisely, this sum lies between 1 and $$n$$ for all $$s > 0$$. Taking the $$s$$-th root:

$$1 \leq \left[\text{sum}\right]^{1/s} \leq n^{1/s}$$

Since $$n^{1/s} \to n^0 = 1$$ as $$s \to \infty$$, by the Squeeze Theorem:

$$\lim_{s \to \infty}\left[\text{sum}\right]^{1/s} = 1$$

Combine results.

$$L = n \times 1 = n$$

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