`lim_(x to pi//2) [x tan x -(pi/2) sec x]`=

Options

A. 1
B. `-1`
C. 0
D. none

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{2}} \left[ x \tan x - \frac{\pi}{2} \sec x \right] \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting \( \tan x \) and \( \sec x \) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \sec x = \frac{1}{\cos x} \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to \frac{\pi}{2}} \left[ x \frac{\sin x}{\cos x} - \frac{\pi}{2} \frac{1}{\cos x} \right] \]

### Step 2: Combine the terms
We can combine the terms over a common denominator:
\[ \lim_{x \to \frac{\pi}{2}} \left[ \frac{x \sin x - \frac{\pi}{2}}{\cos x} \right] \]

### Step 3: Evaluate the limit
Substituting \( x = \frac{\pi}{2} \):
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

This gives us the form \( \frac{0}{0} \), which is indeterminate. Therefore, we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- Derivative of the numerator \( x \sin x - \frac{\pi}{2} \) is:
\[ \frac{d}{dx}(x \sin x) = \sin x + x \cos x \]
- Derivative of the denominator \( \cos x \) is:
\[ \frac{d}{dx}(\cos x) = -\sin x \]

Now we can rewrite the limit:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\sin x + x \cos x}{-\sin x} \]

### Step 5: Substitute \( x = \frac{\pi}{2} \) again
Substituting \( x = \frac{\pi}{2} \):
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1 + \frac{\pi}{2} \cdot 0}{-1} = \frac{1}{-1} = -1 \]

### Final Answer
The limit is:
\[ \lim_{x \to \frac{\pi}{2}} \left[ x \tan x - \frac{\pi}{2} \sec x \right] = -1 \]