If `f (x)= {{:((sin {cos x})/(x- (pi)/(2)) , x ne (pi)/(2)),(1, x= (pi)/(2)):},` where {k} represents the fractional park of k, then:

Options

A. f (x) is continous at `x = (pi)/(2)`
B. `lim _(x to (pi)/2)` f (x) does not exist
C. `lim _(x to (pi)/(2))f (x)` exists, but f is not continuous at `x = (pi)/(2)`
D. `lim _(x to (pi)/(2))f (x)=1`

Answer

To determine the continuity of the function \( f(x) \) at \( x = \frac{\pi}{2} \), we need to check the left-hand limit (LHL) and the right-hand limit (RHL) at that point, and see if they equal \( f\left(\frac{\pi}{2}\right) \).

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

### Step 1: Calculate the Left-Hand Limit (LHL)

We calculate the left-hand limit as \( x \) approaches \( \frac{\pi}{2} \) from the left:
\[ \text{LHL} = \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} \frac{\sin(\cos x)}{x - \frac{\pi}{2}} \]

### Step 2: Substitute \( x = \frac{\pi}{2} - h \)

Let \( h \) be a small positive number, then as \( x \to \frac{\pi}{2}^- \), \( h \to 0^+ \):
\[ \text{LHL} = \lim_{h \to 0^+} \frac{\sin(\cos(\frac{\pi}{2} - h))}{\frac{\pi}{2} - h - \frac{\pi}{2}} = \lim_{h \to 0^+} \frac{\sin(\sin h)}{-h} \]

### Step 3: Simplify the Limit

As \( h \to 0 \), \( \sin h \to h \):
\[ \text{LHL} = \lim_{h \to 0^+} \frac{\sin(\sin h)}{-h} \]

Using the small angle approximation \( \sin x \approx x \):
\[ \text{LHL} = \lim_{h \to 0^+} \frac{\sin(h)}{-h} \]

### Step 4: Apply L'Hôpital's Rule

Since this is an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:
\[ \text{LHL} = \lim_{h \to 0^+} \frac{\cos(h)}{-1} = -\cos(0) = -1 \]

### Step 5: Calculate the Right-Hand Limit (RHL)

Now we calculate the right-hand limit as \( x \) approaches \( \frac{\pi}{2} \) from the right:
\[ \text{RHL} = \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \frac{\sin(\cos x)}{x - \frac{\pi}{2}} \]

### Step 6: Substitute \( x = \frac{\pi}{2} + h \)

Let \( h \) be a small positive number, then as \( x \to \frac{\pi}{2}^+ \), \( h \to 0^+ \):
\[ \text{RHL} = \lim_{h \to 0^+} \frac{\sin(\cos(\frac{\pi}{2} + h))}{\frac{\pi}{2} + h - \frac{\pi}{2}} = \lim_{h \to 0^+} \frac{\sin(-\sin h)}{h} \]

### Step 7: Simplify the Limit

Using the property \( \sin(-x) = -\sin(x) \):
\[ \text{RHL} = \lim_{h \to 0^+} \frac{-\sin(\sin h)}{h} \]

### Step 8: Apply L'Hôpital's Rule Again

This is again an indeterminate form \( \frac{0}{0} \):
\[ \text{RHL} = \lim_{h \to 0^+} \frac{-\cos(\sin h) \cos h}{1} = -\cos(0) = -1 \]

### Step 9: Compare Limits and Function Value

We have:
- \( \text{LHL} = -1 \)
- \( \text{RHL} = -1 \)
- \( f\left(\frac{\pi}{2}\right) = 1 \)

Since \( \text{LHL} \neq f\left(\frac{\pi}{2}\right) \) and \( \text{RHL} \neq f\left(\frac{\pi}{2}\right) \), the function is not continuous at \( x = \frac{\pi}{2} \).

### Conclusion

The function \( f(x) \) is not continuous at \( x = \frac{\pi}{2} \).

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