`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.
`L= lim_(xto0-) f(x)` is equal to

Options

A. `(pi)/(2)`
B. `(pi)/(2sqrt(2))`
C. `(pi)`
D. `(1)/(2sqrt(2))`

Explanation

To find the limit \( L = \lim_{x \to 0^-} f(x) \) where

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

we need to analyze the behavior of the function as \( x \) approaches \( 0 \) from the left.

### Step 1: Understand the fractional part function
The fractional part of \( x \), denoted as \( \{x\} \), is defined as \( x - \lfloor x \rfloor \). As \( x \to 0^- \), \( \lfloor x \rfloor = -1 \), hence:

\[ \{x\} = x - (-1) = x + 1 \]

So, as \( x \to 0^- \), \( \{x\} \to 1 \).

### Step 2: Substitute \(\{x\}\) into \(f(x)\)
Now we can rewrite \( f(x) \):

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

This simplifies to:

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

### Step 3: Simplify the expression
Using the identities \( \sin^{-1}(-x) = -\sin^{-1}(x) \) and \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \):

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

This simplifies to:

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

### Step 4: Find the limit as \( x \to 0^- \)
As \( x \to 0^- \):

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

Thus, \( \pi - \cos^{-1}(x) \to \frac{\pi}{2} \).

Now substituting these values into \( f(x) \):

\[ f(x) \to \frac{0 \cdot \frac{\pi}{2}}{\sqrt{2(1)}} = 0 \]

### Conclusion
Therefore, the limit is:

\[ L = \lim_{x \to 0^-} f(x) = 0 \]