By using L' Hospital's rule, evaluate : `Lim_(x rarr (pi)/(2))` (cos x log tan x).

Answer

To evaluate the limit \( \lim_{x \to \frac{\pi}{2}} \cos x \log \tan x \) using L'Hospital's Rule, we first need to rewrite the expression in a suitable form.

### Step 1: Rewrite the limit
We start with:
\[ \lim_{x \to \frac{\pi}{2}} \cos x \log \tan x \]
As \( x \) approaches \( \frac{\pi}{2} \), \( \cos x \) approaches \( 0 \) and \( \log \tan x \) approaches \( -\infty \) (since \( \tan x \to \infty \)). Therefore, we have a \( 0 \cdot (-\infty) \) form, which we can rewrite as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\log \tan x}{\frac{1}{\cos x}} \]
This gives us the form \( \frac{-\infty}{\infty} \), which is suitable for applying L'Hospital's Rule.

### Step 2: Apply L'Hospital's Rule
Now, we differentiate the numerator and the denominator:
- The derivative of the numerator \( \log \tan x \) is:
\[ \frac{d}{dx}(\log \tan x) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \]
- The derivative of the denominator \( \frac{1}{\cos x} \) is:
\[ \frac{d}{dx}(\frac{1}{\cos x}) = \frac{\sin x}{\cos^2 x} = \tan x \sec x \]

Thus, we can rewrite our limit as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\frac{\sec^2 x}{\tan x}}{\tan x \sec x} \]

### Step 3: Simplify the expression
This simplifies to:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\sec^2 x}{\tan^2 x \sec x} = \lim_{x \to \frac{\pi}{2}} \frac{\sec x}{\tan^2 x} \]

### Step 4: Substitute the limit
Now, we substitute \( x = \frac{\pi}{2} \):
- \( \sec \frac{\pi}{2} = \frac{1}{\cos \frac{\pi}{2}} = \infty \)
- \( \tan \frac{\pi}{2} = \infty \)

Thus, we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\infty}{\infty^2} = \lim_{x \to \frac{\pi}{2}} \frac{1}{\tan^2 x} \]

### Step 5: Evaluate the limit
As \( x \to \frac{\pi}{2} \), \( \tan^2 x \to \infty \), hence:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1}{\tan^2 x} = 0 \]

### Conclusion
Thus, the final result is:
\[ \lim_{x \to \frac{\pi}{2}} \cos x \log \tan x = 0 \]