`lim _(x to (pi)/(2)) (cos x ) ^(cos x )` is :

Options

A. 1
B. 0
C. `1/e`
D. `2/e`

Answer

To find the limit \( \lim_{x \to \frac{\pi}{2}} (\cos x)^{\cos x} \), we will follow these steps:

### Step 1: Set the limit equal to a variable
Let
\[ L = \lim_{x \to \frac{\pi}{2}} (\cos x)^{\cos x} \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we get:
\[ \log L = \lim_{x \to \frac{\pi}{2}} \cos x \cdot \log(\cos x) \]

### Step 3: Analyze the limit
As \( x \to \frac{\pi}{2} \), \( \cos x \to 0 \) and \( \log(\cos x) \to -\infty \). Therefore, we have a \( 0 \cdot (-\infty) \) form, which is indeterminate. We can rewrite this as:
\[ \log L = \lim_{x \to \frac{\pi}{2}} \frac{\log(\cos x)}{\frac{1}{\cos x}} \]
This is now in the form \( \frac{-\infty}{\infty} \), which is suitable for applying L'Hospital's Rule.

### Step 4: Apply L'Hospital's Rule
Differentiating the numerator and the denominator:
- The derivative of \( \log(\cos x) \) is \( -\tan x \).
- The derivative of \( \frac{1}{\cos x} \) is \( \sec x \tan x \).

Applying L'Hospital's Rule gives us:
\[ \log L = \lim_{x \to \frac{\pi}{2}} \frac{-\tan x}{\sec x \tan x} = \lim_{x \to \frac{\pi}{2}} -\cos x \]

### Step 5: Evaluate the limit
As \( x \to \frac{\pi}{2} \), \( -\cos x \to -0 \), which approaches \( 0 \). Thus:
\[ \log L = 0 \]

### Step 6: Solve for L
Exponentiating both sides, we find:
\[ L = e^0 = 1 \]

### Conclusion
Therefore, the limit is:
\[ \lim_{x \to \frac{\pi}{2}} (\cos x)^{\cos x} = 1 \]