1. Rewrite the expression

We need to evaluate

$$L=\lim_{t\to 0}\left(1^{\frac1{\sin^2 t}}+2^{\frac1{\sin^2 t}}+\cdots+n^{\frac1{\sin^2 t}}\right)^{\sin^2 t}.$$

Let

$$m=\frac1{\sin^2 t}.$$

As $t\to 0$, we have $\sin^2 t\to 0^+$, so

$$m\to \infty.$$

Thus the limit becomes

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

2. Recognize the standard form

This is of the form

$$\left(\sum_{k=1}^n k^m\right)^{1/m}$$

as $m\to\infty$.

For very large $m$, the largest term dominates the sum. Among

$$1^m,2^m,\dots,n^m,$$

the largest is $n^m$.

So we expect

$$\left(1^m+2^m+\cdots+n^m\right)^{1/m}\to n.$$

3. Justify using inequalities

Since each term is nonnegative,

$$n^m \le 1^m+2^m+\cdots+n^m \le n\cdot n^m = n^{m+1}.$$

Taking $m$-th root throughout,

$$n \le \left(1^m+2^m+\cdots+n^m\right)^{1/m} \le n^{1+1/m}=n\,n^{1/m}.$$

Now as $m\to\infty$,

$$n^{1/m}\to 1.$$

Hence both lower and upper bounds tend to $n$. By the squeeze theorem,

$$L=n.$$

4. Check options

• A: $\dfrac{n(n+1)}2$ ❌
• B: $n$ ✅
• C: $n^2+n$ ❌
• D: $n^2$ ❌

Therefore, the correct option is B.