If [ ] denotes the greatest integer function, `lim_(x to(pi)/2)(5 sin [cos x])/([cos x]+2)` is

Options

A. 0
B. 1
C. `oo`
D. Does not exist

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{2}} \frac{5 \sin [\cos x]}{[\cos x] + 2} \), where \([\cdot]\) denotes the greatest integer function, we will analyze the behavior of \(\cos x\) as \(x\) approaches \(\frac{\pi}{2}\) from both sides.

### Step-by-Step Solution:

1. **Understand the behavior of \(\cos x\)**:
- As \(x\) approaches \(\frac{\pi}{2}\), \(\cos x\) approaches \(0\).
- Specifically, for \(x\) slightly less than \(\frac{\pi}{2}\) (i.e., \(x \to \frac{\pi}{2}^-\)), \(\cos x\) is positive and approaches \(0\).
- For \(x\) slightly more than \(\frac{\pi}{2}\) (i.e., \(x \to \frac{\pi}{2}^+\)), \(\cos x\) is negative and also approaches \(0\).

2. **Calculate the left-hand limit**:
- For \(x \to \frac{\pi}{2}^-\):
- \(\cos x\) is in the interval \((0, 1)\).
- Therefore, \([\cos x] = 0\).
- The limit becomes:
\[ \lim_{x \to \frac{\pi}{2}^-} \frac{5 \sin [\cos x]}{[\cos x] + 2} = \lim_{x \to \frac{\pi}{2}^-} \frac{5 \sin 0}{0 + 2} = \frac{5 \cdot 0}{2} = 0. \]

3. **Calculate the right-hand limit**:
- For \(x \to \frac{\pi}{2}^+\):
- \(\cos x\) is in the interval \((-1, 0)\).
- Therefore, \([\cos x] = -1\).
- The limit becomes:
\[ \lim_{x \to \frac{\pi}{2}^+} \frac{5 \sin [\cos x]}{[\cos x] + 2} = \lim_{x \to \frac{\pi}{2}^+} \frac{5 \sin (-1)}{-1 + 2} = \frac{5 \sin (-1)}{1} = 5 \sin (-1). \]

4. **Compare the left-hand and right-hand limits**:
- The left-hand limit is \(0\).
- The right-hand limit is \(5 \sin (-1)\), which is not equal to \(0\) (since \(\sin(-1) \neq 0\)).
- Therefore, the left-hand limit does not equal the right-hand limit.

5. **Conclusion**:
- Since the left-hand limit and the right-hand limit are not equal, the limit does not exist.
- Thus, we conclude that:
\[ \lim_{x \to \frac{\pi}{2}} \frac{5 \sin [\cos x]}{[\cos x] + 2} \text{ does not exist.} \]