Question

If \( 0 \leq x \leq \frac{\pi}{2} \), then \[ \lim\limits_{x \to a} \frac{2\cos x - 1}{2\cos x - 1} \] Options: 

• A. \text{does not exist at all points in} \( \left[0, \frac{\pi}{2} \right] \)
• B. \text{ }
• C. \( = -1, \) \text{when} \( a = \frac{\pi}{3} \)
• D. \( = 1, \) \text{when} \( 0 \leq a < \frac{\pi}{3} \)
  \

Hint:

To analyze a limit expression, check where the denominator becomes zero. The function behaves differently on either side of these points.

Answer 1

The Correct Option is D

Step 1: Evaluating the Given Limit
The given function is: \[ f(x) = \frac{2\cos x - 1}{2\cos x - 1}. \] For all values of \( x \) where \( 2\cos x - 1 \neq 0 \), we get: \[ \lim\limits_{x \to a} f(x) = 1. \] Step 2: Identifying the Undefined Points
The denominator becomes zero when: \[ 2\cos a - 1 = 0. \] \[ \cos a = \frac{1}{2}. \] \[ a = \frac{\pi}{3}. \] At \( a = \frac{\pi}{3} \), the function is undefined. Step 3: Determining Where the Limit Exists
For \( 0 \leq a<\frac{\pi}{3} \), the denominator does not become zero, and we have: \[ \lim\limits_{x \to a} f(x) = 1. \] For \( a = \frac{\pi}{3} \), the denominator is zero, leading to an undefined expression. For \( a>\frac{\pi}{3} \), the expression changes sign, yielding: \[ \lim\limits_{x \to a} f(x) = -1. \] Step 4: Conclusion
Thus, the final answer is: \[ \boxed{1, \text{ when } 0 \leq a<\frac{\pi}{3}.} \] \bigskip