`lim_(x to pi//2) (2x -pi)/(cos x)`=

Options

A. 1
B. 2
C. `-2`
D. `0

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{2}} \frac{2x - \pi}{\cos x} \), we can follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{2} \)
First, we substitute \( x = \frac{\pi}{2} \) into the expression:
\[ \frac{2\left(\frac{\pi}{2}\right) - \pi}{\cos\left(\frac{\pi}{2}\right)} = \frac{\pi - \pi}{0} = \frac{0}{0} \]
Since we get an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of the numerator \( 2x - \pi \) is \( 2 \).
- The derivative of the denominator \( \cos x \) is \( -\sin x \).

Thus, we rewrite the limit as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{2}{-\sin x} \]

### Step 3: Substitute \( x = \frac{\pi}{2} \) again
Now we substitute \( x = \frac{\pi}{2} \) into the new expression:
\[ \frac{2}{-\sin\left(\frac{\pi}{2}\right)} = \frac{2}{-1} = -2 \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to \frac{\pi}{2}} \frac{2x - \pi}{\cos x} = -2 \]