The value of `lim_(xrarr0) {1^((1)/(sin^(2)x)+)2^((1)/(sin^(2)x))+3^((1)/(sin^(2)x))+.....+n^((1)/sin^(2)x)}^(sin^2x)`, is

Options

A. `oo`
B. `0`
C. `n`
D. `(n(n+1))/(2)`

Explanation

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{x \to 0} \left( 1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \ldots + n^{\frac{1}{\sin^2 x}} \right)^{\sin^2 x} \]

### Step 1: Rewrite the expression
We can rewrite the expression inside the limit as follows:

\[ S = 1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \ldots + n^{\frac{1}{\sin^2 x}} \]

Thus, we need to evaluate:

\[ \lim_{x \to 0} S^{\sin^2 x} \]

### Step 2: Analyze the limit as \( x \to 0 \)
As \( x \to 0 \), \( \sin^2 x \) approaches 0. Therefore, we need to analyze the behavior of \( S \) as \( x \to 0 \).

### Step 3: Evaluate \( S \)
The term \( k^{\frac{1}{\sin^2 x}} \) for each \( k \) (where \( k = 1, 2, \ldots, n \)) approaches infinity as \( x \to 0 \). Hence, we can express \( S \) as:

\[ S = 1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + \ldots + n^{\frac{1}{\sin^2 x}} \approx 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \ldots + n^{\frac{1}{\sin^2 x}} \]

As \( x \to 0 \), the dominant term in \( S \) will be \( n^{\frac{1}{\sin^2 x}} \).

### Step 4: Simplify \( S \)
Thus, we can approximate:

\[ S \approx n^{\frac{1}{\sin^2 x}} \]

### Step 5: Substitute back into the limit
Now, substituting back into our limit, we have:

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

This simplifies to:

\[ \lim_{x \to 0} n^{\sin^2 x \cdot \frac{1}{\sin^2 x}} = \lim_{x \to 0} n^1 = n \]

### Final Answer
Thus, the value of the limit is:

\[ \boxed{n} \]