To solve the limit problem given by
\[
L = \lim_{x \to 0} \frac{|1 - x + |x||}{|\lambda - x + [x]|}
\]
where \(\lambda \in \mathbb{R} \setminus \{0, 1\}\) and \([x]\) denotes the greatest integer function (G.I.F.), we will analyze the limit from both the left-hand side (LHL) and the right-hand side (RHL) as \(x\) approaches 0.
### Step 1: Analyze the left-hand limit (LHL)
When \(x\) approaches 0 from the left (\(x \to 0^{-}\)), we can substitute \(x = -h\) where \(h \to 0^{+}\). Thus, we have:
\[
LHL = \lim_{h \to 0^{+}} \frac{|1 - (-h) + |-h||}{|\lambda - (-h) + [-h]|}
\]
Calculating the components:
- \(|-h| = h\), so the numerator becomes:
\[
|1 + h + h| = |1 + 2h| = 1 + 2h \quad \text{(as } h \to 0^{+}\text{)}
\]
- For the denominator, \([-h] = -1\) (since \([-h]\) is the greatest integer less than or equal to \(-h\)), thus:
\[
|\lambda + h + (-1)| = |\lambda + h - 1| = |\lambda - 1 + h|
\]
Putting it together, we have:
\[
LHL = \lim_{h \to 0^{+}} \frac{1 + 2h}{|\lambda - 1 + h|}
\]
As \(h \to 0^{+}\), this simplifies to:
\[
LHL = \frac{1}{|\lambda - 1|}
\]
### Step 2: Analyze the right-hand limit (RHL)
Now, consider \(x\) approaching 0 from the right (\(x \to 0^{+}\)). We substitute \(x = h\) where \(h \to 0^{+}\):
\[
RHL = \lim_{h \to 0^{+}} \frac{|1 - h + |h||}{|\lambda - h + [h]|}
\]
Calculating the components:
- \(|h| = h\), so the numerator becomes:
\[
|1 - h + h| = |1| = 1
\]
- For the denominator, \([h] = 0\) (since \(h\) is positive and less than 1), thus:
\[
|\lambda - h + 0| = |\lambda - h|
\]
Putting it together, we have:
\[
RHL = \lim_{h \to 0^{+}} \frac{1}{|\lambda - h|}
\]
As \(h \to 0^{+}\), this simplifies to:
\[
RHL = \frac{1}{|\lambda|}
\]
### Step 3: Set LHL equal to RHL
For the limit \(L\) to exist, we must have:
\[
\frac{1}{|\lambda - 1|} = \frac{1}{|\lambda|}
\]
Cross-multiplying gives:
\[
|\lambda| = |\lambda - 1|
\]
### Step 4: Solve the equation
This absolute value equation can be split into two cases:
1. \(\lambda = \lambda - 1\) which simplifies to \(1 = 0\) (not possible).
2. \(\lambda = -(\lambda - 1)\) which simplifies to \(2\lambda = 1\) or \(\lambda = \frac{1}{2}\).
However, since \(\lambda\) must not be \(0\) or \(1\), we check if \(\lambda = \frac{1}{2}\) is valid. It is valid since it is not in the excluded set.
### Conclusion
Thus, the limit \(L\) is given by:
\[
L = \frac{1}{|\lambda - 1|} = \frac{1}{|\frac{1}{2} - 1|} = \frac{1}{\frac{1}{2}} = 2
\]
So, the final answer is:
\[
\boxed{2}
\]
