Question

The value of \( \lim_{x \to \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right) \) is:

• A. 1
• B. \( -1 \)
• C. 0
• D. \( \pi \)

Hint:

When dealing with indeterminate forms like \( \frac{0}{0} \), you can apply L'Hopital's Rule to differentiate the numerator and denominator and evaluate the limit.

Answer 1

The Correct Option is B

We are tasked with finding the limit: \[ \lim_{x \to \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right) \] First, note that \( \cos \left( \frac{\pi}{2} \right) = 0 \), so both the numerator and denominator approach 0 as \( x \to \frac{\pi}{2} \). This gives us an indeterminate form \( \frac{0}{0} \), and we can apply L'Hopital’s Rule to evaluate the limit.
Taking the derivative of the numerator and denominator:
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( x - \frac{\pi}{2} \) is simply 1.
Thus, applying L'Hopital’s Rule: \[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{x - \frac{\pi}{2}} = \lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{1} \] Since \( \sin \left( \frac{\pi}{2} \right) = 1 \), we get: \[ \lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{1} = -1 \] Thus, the value of the limit is \( -1 \), and the correct answer is option (B).