If F(x) = `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `lim_(xto pi//2)f(x)` is

Options

A. `-1`
B. 1
C. non-existant
D. none of these

Explanation

To solve the limit \( \lim_{x \to \frac{\pi}{2}} F(x) \) where

\[ F(x) = \begin{cases} \frac{\sin(\{ \cos x \})}{x - \frac{\pi}{2}}, & x \neq \frac{\pi}{2} \\ 1, & x = \frac{\pi}{2} \end{cases} \]

we need to evaluate the limit as \( x \) approaches \( \frac{\pi}{2} \) from both sides.

### Step 1: Evaluate the Right-Hand Limit (RHL)

We start by calculating the right-hand limit as \( x \) approaches \( \frac{\pi}{2} \) from the positive side:

\[ \lim_{x \to \frac{\pi}{2}^+} F(x) = \lim_{x \to \frac{\pi}{2}^+} \frac{\sin(\{ \cos x \})}{x - \frac{\pi}{2}} \]

### Step 2: Determine \( \cos x \) as \( x \to \frac{\pi}{2}^+ \)

As \( x \) approaches \( \frac{\pi}{2} \) from the right, \( \cos x \) approaches \( 0 \) from the negative side (since \( \cos \) is decreasing in this interval):

\[ \cos\left(\frac{\pi}{2}^+\right) \to 0^- \]

### Step 3: Find the Fractional Part

The fractional part function \( \{ \cos x \} \) is defined as \( \cos x - \lfloor \cos x \rfloor \). Since \( \cos x \) approaches \( 0^- \), we have:

\[ \{ \cos x \} = 1 + \cos x \quad \text{(because } \lfloor \cos x \rfloor = -1 \text{ for } \cos x < 0\text{)} \]

Thus, as \( x \to \frac{\pi}{2}^+ \):

\[ \{ \cos x \} \to 1 + 0^- = 1 \]

### Step 4: Substitute into the Limit

Now substituting back into the limit:

\[ \lim_{x \to \frac{\pi}{2}^+} F(x) = \lim_{x \to \frac{\pi}{2}^+} \frac{\sin(1)}{x - \frac{\pi}{2}} \]

As \( x \to \frac{\pi}{2}^+ \), \( x - \frac{\pi}{2} \to 0^+ \). Therefore, the limit becomes:

\[ \lim_{x \to \frac{\pi}{2}^+} \frac{\sin(1)}{x - \frac{\pi}{2}} \to \infty \]

### Step 5: Evaluate the Left-Hand Limit (LHL)

Now, we calculate the left-hand limit as \( x \) approaches \( \frac{\pi}{2} \) from the negative side:

\[ \lim_{x \to \frac{\pi}{2}^-} F(x) = \lim_{x \to \frac{\pi}{2}^-} \frac{\sin(\{ \cos x \})}{x - \frac{\pi}{2}} \]

As \( x \) approaches \( \frac{\pi}{2} \) from the left, \( \cos x \) approaches \( 0 \) from the positive side:

\[ \cos\left(\frac{\pi}{2}^-\right) \to 0^+ \]

Thus, the fractional part becomes:

\[ \{ \cos x \} = \cos x \quad \text{(because } \lfloor \cos x \rfloor = 0 \text{ for } \cos x \geq 0\text{)} \]

So we have:

\[ \lim_{x \to \frac{\pi}{2}^-} F(x) = \lim_{x \to \frac{\pi}{2}^-} \frac{\sin(\cos x)}{x - \frac{\pi}{2}} \]

As \( x \to \frac{\pi}{2}^- \), \( \cos x \to 0^+ \) and thus \( \sin(\cos x) \to \sin(0) = 0 \).

### Step 6: Evaluate the Limit

Now substituting back into the limit:

\[ \lim_{x \to \frac{\pi}{2}^-} \frac{\sin(\cos x)}{x - \frac{\pi}{2}} \to \frac{0}{0^-} \to 0 \]

### Conclusion

Since the right-hand limit approaches \( \infty \) and the left-hand limit approaches \( 0 \), we conclude that:

\[ \lim_{x \to \frac{\pi}{2}} F(x) \text{ does not exist.} \]

### Final Answer

The limit \( \lim_{x \to \frac{\pi}{2}} F(x) \) does not exist.

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