`Let f(x)=(sin^(-1)(1-{x})xxcos^(-1)(1-{x}))/(sqrt(2{x})xx(1-{x}))`, where `{x}` denotes the fractional part of x.
`R=lim_(xto0+) f(x)` is equal to

Explanation

To solve the limit \( R = \lim_{x \to 0^+} f(x) \) where

\[ f(x) = \frac{\sin^{-1}(1 - \{x\}) \cdot \cos^{-1}(1 - \{x\})}{\sqrt{2\{x\}} \cdot (1 - \{x\})} \]

we start by analyzing the fractional part of \( x \), denoted as \( \{x\). The fractional part can be expressed as:

\[ \{x\} = x - \lfloor x \rfloor \]

As \( x \) approaches \( 0^+ \), the greatest integer function \( \lfloor x \rfloor \) will be \( 0 \). Therefore, we have:

\[ \{x\} = x \]

Substituting this into \( f(x) \):

\[ f(x) = \frac{\sin^{-1}(1 - x) \cdot \cos^{-1}(1 - x)}{\sqrt{2x} \cdot (1 - x)} \]

Next, we need to evaluate the limit as \( x \) approaches \( 0^+ \):

\[ R = \lim_{x \to 0^+} \frac{\sin^{-1}(1 - x) \cdot \cos^{-1}(1 - x)}{\sqrt{2x} \cdot (1 - x)} \]

As \( x \to 0^+ \):

- \( \sin^{-1}(1 - x) \to \sin^{-1}(1) = \frac{\pi}{2} \)
- \( \cos^{-1}(1 - x) \to \cos^{-1}(1) = 0 \)

Thus, the limit takes the form \( \frac{\frac{\pi}{2} \cdot 0}{\sqrt{2 \cdot 0} \cdot 1} \), which is \( \frac{0}{0} \), an indeterminate form. We can apply L'Hôpital's Rule.

To apply L'Hôpital's Rule, we differentiate the numerator and denominator separately:

1. **Differentiate the numerator**:
\[ \frac{d}{dx} \left( \sin^{-1}(1 - x) \cdot \cos^{-1}(1 - x) \right) \]

Using the product rule:
\[ u = \sin^{-1}(1 - x), \quad v = \cos^{-1}(1 - x) \]
\[ \frac{du}{dx} = -\frac{1}{\sqrt{1 - (1 - x)^2}} = -\frac{1}{\sqrt{2x - x^2}} \]
\[ \frac{dv}{dx} = -\frac{1}{\sqrt{1 - (1 - x)^2}} = -\frac{1}{\sqrt{2x - x^2}} \]

Thus,
\[ \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

2. **Differentiate the denominator**:
\[ \frac{d}{dx} \left( \sqrt{2x} \cdot (1 - x) \right) \]
Using the product rule:
\[ \frac{d}{dx}(\sqrt{2x}) = \frac{1}{\sqrt{2x}} \cdot 1 = \frac{1}{\sqrt{2x}}, \quad \frac{d}{dx}(1 - x) = -1 \]
So,
\[ \frac{d}{dx}(\sqrt{2x}(1 - x)) = \sqrt{2x} \cdot (-1) + (1 - x) \cdot \frac{1}{\sqrt{2x}} = -\sqrt{2x} + \frac{1 - x}{\sqrt{2x}} \]

Now we can apply L'Hôpital's Rule:

\[ R = \lim_{x \to 0^+} \frac{\text{Numerator's derivative}}{\text{Denominator's derivative}} \]

After applying L'Hôpital's Rule and simplifying, we will find the limit as \( x \to 0^+ \).

Finally, substituting \( x = 0 \) into the simplified expression will yield the final result.

### Final Result:
\[ R = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} \]