The value f `underset(x to pi/2)lim [1^(1//cos^(2)x)+2^(1//cos^(2) x)+....+n^(1//cos^(2) x)]^(cos^(2)x)` is

Options

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

Answer

To solve the limit problem given, we need to evaluate the following limit:

\[ \lim_{x \to \frac{\pi}{2}} \left[1^{\frac{1}{\cos^2 x}} + 2^{\frac{1}{\cos^2 x}} + \ldots + n^{\frac{1}{\cos^2 x}}\right]^{\cos^2 x} \]

### Step 1: Rewrite the limit expression

First, we can rewrite the expression using the fact that \(\frac{1}{\cos^2 x} = \sec^2 x\):

\[ \lim_{x \to \frac{\pi}{2}} \left[1^{\sec^2 x} + 2^{\sec^2 x} + \ldots + n^{\sec^2 x}\right]^{\cos^2 x} \]

### Step 2: Factor out \(n^{\sec^2 x}\)

Next, we can factor out \(n^{\sec^2 x}\) from the sum:

\[ = \lim_{x \to \frac{\pi}{2}} n^{\sec^2 x} \left[\left(\frac{1}{n}\right)^{\sec^2 x} + \left(\frac{2}{n}\right)^{\sec^2 x} + \ldots + 1^{\sec^2 x}\right]^{\cos^2 x} \]

### Step 3: Simplify the expression inside the limit

Now, we can express the limit as:

\[ = \lim_{x \to \frac{\pi}{2}} n^{\sec^2 x \cdot \cos^2 x} \left[\left(\frac{1}{n}\right)^{\sec^2 x} + \left(\frac{2}{n}\right)^{\sec^2 x} + \ldots + 1^{\sec^2 x}\right]^{\cos^2 x} \]

Using the identity \(\sec^2 x \cdot \cos^2 x = 1\):

\[ = \lim_{x \to \frac{\pi}{2}} n \left[\left(\frac{1}{n}\right)^{\sec^2 x} + \left(\frac{2}{n}\right)^{\sec^2 x} + \ldots + 1^{\sec^2 x}\right]^{\cos^2 x} \]

### Step 4: Evaluate the limit as \(x \to \frac{\pi}{2}\)

As \(x\) approaches \(\frac{\pi}{2}\), \(\sec^2 x\) approaches infinity. Therefore, terms of the form \(\left(\frac{k}{n}\right)^{\sec^2 x}\) for \(k < n\) will approach 0, and the term for \(k = n\) will approach 1:

\[ \lim_{x \to \frac{\pi}{2}} \left[\left(\frac{1}{n}\right)^{\sec^2 x} + \left(\frac{2}{n}\right)^{\sec^2 x} + \ldots + 1^{\sec^2 x}\right]^{\cos^2 x} = 1^{\cos^2 x} = 1 \]

### Step 5: Final result

Thus, the limit simplifies to:

\[ \lim_{x \to \frac{\pi}{2}} n \cdot 1 = n \]

So, the final answer is:

\[ \boxed{n} \]