`lim_(trarr0)(1^(1/sin^2t)+2^(1/sin^2t)+..n^(1/sin^2t))^(sin^2t)`

Options

A. `n^2`
B. `n^(2) +n`
C. n
D. `(n(n+1))/2`

Answer

To solve the limit problem

\[ \lim_{t \to 0} \left(1^{\frac{1}{\sin^2 t}} + 2^{\frac{1}{\sin^2 t}} + \ldots + n^{\frac{1}{\sin^2 t}}\right)^{\sin^2 t}, \]

we will break it down step by step.

### Step 1: Rewrite the expression
We start by rewriting the expression inside the limit:

\[ S(t) = 1^{\frac{1}{\sin^2 t}} + 2^{\frac{1}{\sin^2 t}} + \ldots + n^{\frac{1}{\sin^2 t}}. \]

### Step 2: Analyze each term as \( t \to 0 \)
As \( t \to 0 \), \( \sin^2 t \) approaches 0. Therefore, each term \( k^{\frac{1}{\sin^2 t}} \) (for \( k = 1, 2, \ldots, n \)) behaves as follows:

- For \( k = 1 \): \( 1^{\frac{1}{\sin^2 t}} = 1 \).
- For \( k > 1 \): \( k^{\frac{1}{\sin^2 t}} \) approaches \( \infty \) since \( k > 1 \) and \( \frac{1}{\sin^2 t} \to \infty \).

Thus, \( S(t) \) can be approximated as:

\[ S(t) \approx 1 + \infty + \infty + \ldots + \infty = \infty. \]

### Step 3: Rewrite the limit
Now, we substitute this back into our limit:

\[ \lim_{t \to 0} S(t)^{\sin^2 t} = \lim_{t \to 0} \infty^{\sin^2 t}. \]

### Step 4: Analyze \( \infty^{\sin^2 t} \)
As \( t \to 0 \), \( \sin^2 t \to 0 \). The expression \( \infty^{\sin^2 t} \) can be interpreted as \( e^{\sin^2 t \cdot \ln(\infty)} \). Since \( \ln(\infty) \) is undefined, we need to consider the limit more carefully.

### Step 5: Apply the limit properties
We can express the limit in terms of \( n \):

\[ S(t) = \sum_{k=1}^{n} k^{\frac{1}{\sin^2 t}} \approx n^{\frac{1}{\sin^2 t}} \text{ (dominant term as } t \to 0). \]

Thus, we can rewrite:

\[ \lim_{t \to 0} S(t)^{\sin^2 t} \approx \lim_{t \to 0} \left(n^{\frac{1}{\sin^2 t}}\right)^{\sin^2 t} = \lim_{t \to 0} n^{\sin^2 t / \sin^2 t} = n^{1} = n. \]

### Final Answer
Therefore, the final answer is:

\[ \lim_{t \to 0} \left(1^{\frac{1}{\sin^2 t}} + 2^{\frac{1}{\sin^2 t}} + \ldots + n^{\frac{1}{\sin^2 t}}\right)^{\sin^2 t} = n. \]