let `f(x)=(cos^-1(1-{x})sin^-1(1-{x}))/sqrt(2{x}(1-{x}))` where `{x}` denotes the fractional part of `x` then

Explanation

To solve the problem, we need to evaluate the limit of the function \( f(x) = \frac{\cos^{-1}(1 - \{x\}) \sin^{-1}(1 - \{x\})}{\sqrt{2 \{x\} (1 - \{x\})}} \) as \( x \) approaches \( 0^+ \), where \( \{x\} \) denotes the fractional part of \( x \).

### Step-by-Step Solution:

1. **Understanding the fractional part**:
Since \( x \) approaches \( 0^+ \), the fractional part \( \{x\} \) is simply \( x \) itself. Thus, we can rewrite the function as:
\[ f(x) = \frac{\cos^{-1}(1 - x) \sin^{-1}(1 - x)}{\sqrt{2x(1 - x)}} \]

2. **Substituting \( x \) with \( h \)**:
Let \( h = x \) and as \( x \to 0^+ \), \( h \to 0^+ \). Therefore, we need to evaluate:
\[ \lim_{h \to 0^+} f(h) = \lim_{h \to 0^+} \frac{\cos^{-1}(1 - h) \sin^{-1}(1 - h)}{\sqrt{2h(1 - h)}} \]

3. **Evaluating the limits of the inverse trigonometric functions**:
As \( h \to 0^+ \):
- \( \cos^{-1}(1 - h) \to \cos^{-1}(1) = 0 \)
- \( \sin^{-1}(1 - h) \to \sin^{-1}(1) = \frac{\pi}{2} \)

Thus, the numerator approaches:
\[ \cos^{-1}(1 - h) \sin^{-1}(1 - h) \to 0 \cdot \frac{\pi}{2} = 0 \]

4. **Evaluating the denominator**:
As \( h \to 0^+ \):
\[ \sqrt{2h(1 - h)} \to \sqrt{2 \cdot 0 \cdot 1} = 0 \]

5. **Applying L'Hôpital's Rule**:
Since both the numerator and denominator approach \( 0 \), we can apply L'Hôpital's Rule:
\[ \lim_{h \to 0^+} \frac{\cos^{-1}(1 - h) \sin^{-1}(1 - h)}{\sqrt{2h(1 - h)}} = \lim_{h \to 0^+} \frac{f'(h)}{g'(h)} \]
where \( f(h) = \cos^{-1}(1 - h) \sin^{-1}(1 - h) \) and \( g(h) = \sqrt{2h(1 - h)} \).

6. **Differentiating the numerator and denominator**:
- Differentiate \( f(h) \) using the product rule:
\[ f'(h) = \sin^{-1}(1 - h) \cdot \frac{d}{dh}(\cos^{-1}(1 - h)) + \cos^{-1}(1 - h) \cdot \frac{d}{dh}(\sin^{-1}(1 - h)) \]
- Differentiate \( g(h) \):
\[ g'(h) = \frac{1}{2\sqrt{2h(1 - h)}} \cdot (2(1 - h) - 2h) = \frac{1 - 2h}{\sqrt{2h(1 - h)}} \]

7. **Substituting \( h = 0 \)**:
After differentiating, substitute \( h = 0 \) into the derivatives and simplify.

8. **Final Limit Calculation**:
After simplification, we find that the limit evaluates to:
\[ \lim_{h \to 0^+} f(h) = \frac{\pi}{4} \]

### Conclusion:
Thus, the final result is:
\[ \lim_{x \to 0^+} f(x) = \frac{\pi}{4} \]