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Step-by-Step Solution
Step 1: Understand the expression and the limit
We are given the limit:
$$
\lim_{x \to 0} \left| \frac{1 - x + |x|}{\lambda - x + [x]} \right| = L
$$
where $[\,t\,]$ denotes the greatest integer less than or equal to $t$. We also know that $\lambda \in \mathbb{R} \setminus \{0, 1\}$.
Step 2: Evaluate the left-hand limit (LHL) as $x \to 0^-$
When $x$ is just slightly less than $0$ (negative), we set $x = h$ with $h \to 0^-$.
Numerator: $1 - x + |x| = 1 - h + (-h) = 1 - 2h.$
Denominator: Since $h$ is slightly negative, $[h] = -1$. Thus, $\lambda - x + [x] = \lambda - h - 1.$
As $h \to 0^-$, $1 - 2h \to 1$ and $\lambda - h - 1 \to \lambda - 1.$ So the left-hand limit becomes:
$$
\text{LHL} = \lim_{h \to 0^-}
\left| \frac{1 - 2h}{\lambda - h - 1} \right|
= \left| \frac{1}{\lambda - 1} \right|.
$$
Step 3: Evaluate the right-hand limit (RHL) as $x \to 0^+$
When $x$ is just slightly greater than $0$ (positive), we set $x = h$ with $h \to 0^+.$
Numerator: $1 - x + |x| = 1 - h + h = 1.$
Denominator: For $h$ slightly positive, $[h] = 0.$ Hence, $\lambda - x + [x] = \lambda - h + 0 \approx \lambda.$
As $h \to 0^+$, the expression becomes:
$$
\text{RHL} = \lim_{h \to 0^+}
\left| \frac{1}{\lambda - h + 0} \right|
= \left| \frac{1}{\lambda} \right|.
$$
Step 4: Impose the condition for the limit to exist
For the limit to exist at $x = 0,$ the left-hand limit must equal the right-hand limit. Therefore,
$$
\left| \frac{1}{\lambda - 1} \right| = \left| \frac{1}{\lambda} \right|.
$$
This simplifies to:
$$
\left| \lambda - 1 \right| = \left| \lambda \right|.
$$
Step 5: Solve the equation for $\lambda$
The equation $\left| \lambda - 1 \right| = \left| \lambda \right|$ can hold in two main cases:
Case 1: $\lambda - 1 = \lambda,$ which is impossible, or
Case 2: $\lambda - 1 = -\lambda \quad \Longrightarrow \quad 2\lambda = 1 \quad \Longrightarrow \quad \lambda = \frac{1}{2}.$
Since $\lambda \neq 0$ and $\lambda \neq 1,$ the valid solution is $\lambda = \frac{1}{2}.$
Step 6: Find the value of $L$
Once $\lambda = \frac{1}{2},$ we substitute back into either $\left| \frac{1}{\lambda - 1} \right|$ or $\left| \frac{1}{\lambda} \right|$ to find $L.$ Using $\left| \frac{1}{\lambda} \right|,$ we get:
$$
L = \left| \frac{1}{\tfrac{1}{2}} \right| = 2.
$$
Final Answer
$L = 2$
