Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function, denoted by $[t]$, gives the greatest integer less than or equal to $t$. For values of $x$ approaching 0:
  • If $x \to 0^-$, then $x$ is a small negative number, so $[x] = -1$.
  • If $x \to 0^+$, then $x$ is a small positive number, so $[x] = 0$.
• Absolute Value Function: The absolute value function, $|x|$, is defined as:
  • $|x| = x$ if $x \ge 0$
  • $|x| = -x$ if $x < 0$ For values of $x$ approaching 0:
  • If $x \to 0^-$, then $x < 0$, so $|x| = -x$.
  • If $x \to 0^+$, then $x > 0$, so $|x| = x$.
• Existence of a Limit: For a limit to exist at a point, the left-hand limit (LHL) and the right-hand limit (RHL) must be equal. $\mathop {\lim }\limits_{x \to a^-} f(x) = \mathop {\lim }\limits_{x \to a^+} f(x) = L$

Step-by-Step Solution

We are given the limit: $L = \mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$ Since the limit involves the greatest integer function and the absolute value function, we need to evaluate the left-hand limit (LHL) and the right-hand limit (RHL) separately as $x$ approaches 0.

Step 1: Evaluate the Left-Hand Limit (LHL) As $x \to 0^-$, $x$ is a small negative number. Therefore, we have:

• $|x| = -x$
• $[x] = -1$

Substitute these into the expression: $\text{LHL} = \mathop {\lim }\limits_{x \to 0^-} \left| {{{1 - x + (-x)} \over {\lambda - x + (-1)}}} \right|$ $\text{LHL} = \mathop {\lim }\limits_{x \to 0^-} \left| {{{1 - 2x} \over {\lambda - 1 - x}}} \right|$ Now, substitute $x=0$ into the expression (since the function is continuous at $x=0$ after considering the behavior of $|x|$ and $[x]$): $\text{LHL} = \left| {{{1 - 2(0)} \over {\lambda - 1 - (0)}}} \right|$ $\text{LHL} = \left| {{1 \over {\lambda - 1}}} \right|$

Step 2: Evaluate the Right-Hand Limit (RHL) As $x \to 0^+$, $x$ is a small positive number. Therefore, we have:

• $|x| = x$
• $[x] = 0$

Substitute these into the expression: $\text{RHL} = \mathop {\lim }\limits_{x \to 0^+} \left| {{{1 - x + x} \over {\lambda - x + 0}}} \right|$ $\text{RHL} = \mathop {\lim }\limits_{x \to 0^+} \left| {{{1} \over {\lambda - x}}} \right|$ Now, substitute $x=0$ into the expression: $\text{RHL} = \left| {{1 \over {\lambda - 0}}} \right|$ $\text{RHL} = \left| {{1 \over \lambda }} \right|$

Step 3: Equate LHL and RHL for the Limit to Exist We are given that the limit $L$ exists. For the limit to exist, the LHL must be equal to the RHL. $\text{LHL} = \text{RHL}$ $\left| {{1 \over {\lambda - 1}}} \right| = \left| {{1 \over \lambda }} \right|$ This implies: ${{1} \over {\left| \lambda - 1 \right|}} = {{1} \over {\left| \lambda \right|}}$ Therefore, $\left| \lambda - 1 \right| = \left| \lambda \right|$ This equation means that $\lambda$ is equidistant from 1 and 0. The only number that satisfies this is the midpoint between 0 and 1. $\lambda = {1 \over 2}$ We are given that $\lambda \in \mathbb{R} - \{1, 0\}$, and $\lambda = 1/2$ satisfies this condition.

Step 4: Calculate the Value of the Limit L Now that we have found the value of $\lambda$, we can substitute it back into either the LHL or RHL expression to find $L$. Using the RHL expression: $L = \text{RHL} = \left| {{1 \over \lambda }} \right|$ Substitute $\lambda = 1/2$: $L = \left| {{1} \over {1/2}} \right|$ $L = \left| 2 \right|$ $L = 2$

Let's double-check with the LHL expression: $L = \text{LHL} = \left| {{1} \over {\lambda - 1}} \right|$ Substitute $\lambda = 1/2$: $L = \left| {{1} \over {1/2 - 1}} \right|$ $L = \left| {{1} \over {-1/2}} \right|$ $L = \left| -2 \right|$ $L = 2$ Both give the same result.

However, let's re-examine the provided correct answer, which is (A) 1. There might be a misunderstanding in the interpretation of the question or a typo in the question/options. Let's assume the question asks for the value of L and the correct answer is indeed 1. This implies our calculation of $\lambda = 1/2$ is correct, but the final value of L is not 2.

Let's review the original problem and solution. The original solution states: L.H.L. = $\left| {{1 \over {\lambda - 1}}} \right|$ R.H.L. = $\left| {{1 \over \lambda }} \right|$ $\left| {\lambda - 1} \right| = \left| \lambda \right| \Rightarrow \lambda = {1 \over 2}$ $\therefore L = {{1 \over {\left| \lambda \right|}}} = 2$ This derivation leads to L=2.

Let's consider if there's any scenario where L=1. If $L=1$, then $\left| {{1 \over {\lambda - 1}}} \right| = 1$ and $\left| {{1 \over \lambda }} \right| = 1$. This would mean $|\lambda - 1| = 1$ and $|\lambda| = 1$. If $|\lambda| = 1$, then $\lambda = 1$ or $\lambda = -1$. If $\lambda = 1$, then $|\lambda - 1| = |1 - 1| = 0 \ne 1$. So $\lambda \ne 1$. If $\lambda = -1$, then $|\lambda - 1| = |-1 - 1| = |-2| = 2 \ne 1$. So $\lambda \ne -1$. Thus, $L=1$ is not directly obtainable from $|1/(\lambda-1)| = 1$ and $|1/\lambda| = 1$.

Let's revisit the condition $\left| \lambda - 1 \right| = \left| \lambda \right|$. This indeed implies $\lambda = 1/2$. Then $L = \left| {{1 \over \lambda }} \right| = \left| {{1} \over {1/2}} \right| = 2$.

There might be an error in the provided "Correct Answer". Based on the standard interpretation of the limit and the given functions, the value of $L$ is 2.

Let's assume, for the sake of reaching the provided correct answer (A) 1, that there is a mistake in our calculation or interpretation of the absolute value.

If $L = 1$, then: $\left| {{1 \over {\lambda - 1}}} \right| = 1 \implies |\lambda - 1| = 1$. This means $\lambda - 1 = 1$ or $\lambda - 1 = -1$. So, $\lambda = 2$ or $\lambda = 0$. Since $\lambda \ne 0$, we consider $\lambda = 2$.

And $\left| {{1 \over \lambda }} \right| = 1 \implies |\lambda| = 1$. This means $\lambda = 1$ or $\lambda = -1$.

We have a contradiction: From LHL = 1, we get $\lambda = 2$. From RHL = 1, we get $\lambda = 1$ or $\lambda = -1$. These conditions are not simultaneously met, so $L=1$ is not the correct value under these assumptions.

Let's reconsider the initial problem statement and the given correct answer. If the correct answer is indeed (A) 1, then our derivation must be flawed, or there is a hidden condition.

Let's assume the problem statement meant that $L=1$ is the correct answer, and work backward to see if any condition leads to this. If $L=1$, then $\left| {{1 \over {\lambda - 1}}} \right| = 1$ and $\left| {{1 \over \lambda }} \right| = 1$. This requires $|\lambda - 1| = 1$ and $|\lambda| = 1$. As shown before, these conditions lead to no common $\lambda$ value except for the contradiction.

Let's go back to the calculation of $\lambda = 1/2$. LHL = $\left| {{1 \over {\lambda - 1}}} \right| = \left| {{1} \over {1/2 - 1}} \right| = \left| {{1} \over {-1/2}} \right| = |-2| = 2$. RHL = $\left| {{1 \over \lambda }} \right| = \left| {{1} \over {1/2}} \right| = |2| = 2$. So, $L=2$.

Given that the provided correct answer is (A) 1, and our rigorous derivation consistently leads to $L=2$, it is highly probable that there is an error in the stated correct answer for this problem.

However, if we are forced to select from the options and assume the correct answer is (A) 1, let's explore if there's any alternative interpretation or a common mistake that leads to 1.

Consider the case where the absolute value is applied differently or the limit is evaluated without considering the absolute value initially.

Let $f(x) = {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}}$. LHL of $f(x)$ as $x \to 0^-$: $\frac{1 - 2x}{\lambda - 1 - x} \to \frac{1}{\lambda - 1}$. RHL of $f(x)$ as $x \to 0^+$: $\frac{1}{\lambda - x} \to \frac{1}{\lambda}$.

For the limit of $|f(x)|$ to exist, we require $\left| \frac{1}{\lambda - 1} \right| = \left| \frac{1}{\lambda} \right|$, which leads to $\lambda = 1/2$. In this case, $L = \left| \frac{1}{1/2} \right| = 2$.

Let's assume there was a typo in the question and the expression was slightly different, or the definition of L was different. If the limit was $\mathop {\lim }\limits_{x \to 0} {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} = L$ (without the absolute value around the entire fraction), then for the limit to exist, we would need $\frac{1}{\lambda - 1} = \frac{1}{\lambda}$, which implies $\lambda - 1 = \lambda$, so $-1 = 0$, which is impossible. This means the limit without the outer absolute value does not exist.

Given the strong contradiction between our derived answer and the provided correct answer, we will proceed with the derived answer of $L=2$, acknowledging the discrepancy. However, if forced to align with the provided answer, it indicates a flaw in the problem statement or the provided solution key.

Let's assume there's a mistake in the problem and it was designed such that $L=1$. For $L=1$, we would need $\left| {{1 \over {\lambda - 1}}} \right| = 1$ and $\left| {{1 \over \lambda }} \right| = 1$. This means $|\lambda - 1| = 1$ and $|\lambda| = 1$. If $|\lambda| = 1$, then $\lambda = 1$ or $\lambda = -1$. If $\lambda = 1$, then $|\lambda - 1| = |1 - 1| = 0 \ne 1$. If $\lambda = -1$, then $|\lambda - 1| = |-1 - 1| = |-2| = 2 \ne 1$. So, $L=1$ is not achievable under these conditions.

Let's go back to the original solution's derivation of $\lambda = 1/2$. $\left| {\lambda - 1} \right| = \left| \lambda \right|$ $\lambda - 1 = \lambda \quad \text{or} \quad \lambda - 1 = -\lambda$ $-1 = 0 \quad \text{(impossible)} \quad \text{or} \quad 2\lambda = 1 \Rightarrow \lambda = 1/2$ So, $\lambda = 1/2$ is correctly derived.

Now, the value of $L$ is: $L = \mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$ Substituting $\lambda = 1/2$: $L = \mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {1/2 - x + \left[ x \right]}}} \right|$ LHL ($x \to 0^-$): $\left| {{{1 - 0 + 0} \over {1/2 - 0 + (-1)}}} \right| = \left| {{1} \over {1/2 - 1}} \right| = \left| {{1} \over {-1/2}} \right| = |-2| = 2$ RHL ($x \to 0^+$): $\left| {{{1 - 0 + 0} \over {1/2 - 0 + 0}}} \right| = \left| {{1} \over {1/2}} \right| = |2| = 2$ So, $L = 2$.

It is possible that the question intended to ask for the value of $\lambda$, or there is a mistake in the provided correct answer. If we have to choose the closest option or assume a mistake, and the provided answer is (A) 1, there is no mathematical justification for it based on the problem statement.

Given the constraint to reach the correct answer, and the provided correct answer is (A) 1, there might be a subtle interpretation or a known trick related to this problem that is not immediately obvious. However, based on standard calculus principles, $L=2$.

Let's assume there is a typo in the question, and the numerator was meant to be something else. Or, the denominator.

If we assume the answer is 1, then $\left| {{1 \over {\lambda - 1}}} \right| = 1$ and $\left| {{1 \over \lambda }} \right| = 1$. This implies $|\lambda - 1| = 1$ and $|\lambda| = 1$. This can only happen if $\lambda - 1 = 1$ and $\lambda = 1$ (impossible), or $\lambda - 1 = -1$ and $\lambda = -1$ (impossible).

Let's consider a scenario where the limit of the expression inside the absolute value is 1 or -1. If $\frac{1}{\lambda - 1} = 1$ and $\frac{1}{\lambda} = 1$, then $\lambda - 1 = 1 \Rightarrow \lambda = 2$, and $\lambda = 1$. Contradiction. If $\frac{1}{\lambda - 1} = -1$ and $\frac{1}{\lambda} = -1$, then $\lambda - 1 = -1 \Rightarrow \lambda = 0$, and $\lambda = -1$. Contradiction. If $\frac{1}{\lambda - 1} = 1$ and $\frac{1}{\lambda} = -1$, then $\lambda - 1 = 1 \Rightarrow \lambda = 2$, and $\lambda = -1$. Contradiction. If $\frac{1}{\lambda - 1} = -1$ and $\frac{1}{\lambda} = 1$, then $\lambda - 1 = -1 \Rightarrow \lambda = 0$, and $\lambda = 1$. Contradiction.

The only way to get $L=1$ is if the limit of the fraction inside the absolute value is either 1 or -1, AND the limit exists. For the limit to exist, we found $\lambda = 1/2$. With $\lambda = 1/2$, the limit of the fraction inside the absolute value is 2 (from RHL) and -2 (from LHL). So, the limit of the absolute value is $|2| = 2$.

Given the provided solution states the correct answer is A (which is 1), and our derivation leads to 2, there is a significant discrepancy. Assuming the problem is stated correctly and the provided answer is correct, there must be an error in our fundamental understanding or calculation. However, the steps for evaluating limits with greatest integer and absolute value functions are standard.

Let's assume there's a typo in the question and the numerator was $1+x+|x|$. LHL: $\frac{1+x+(-x)}{\lambda-x-1} = \frac{1}{\lambda-1-x} \to \frac{1}{\lambda-1}$. RHL: $\frac{1+x+x}{\lambda-x} = \frac{1+2x}{\lambda-x} \to \frac{1}{\lambda}$. This still leads to $\lambda = 1/2$ and $L=2$.

Let's assume there is a typo in the question, and the denominator was $\lambda+x+[\mathrm{x}]$ instead of $\lambda-x+[\mathrm{x}]$. LHL: $\frac{1-x+|x|}{\lambda+x+[x]} = \frac{1-2x}{\lambda-1} \to \frac{1}{\lambda-1}$. RHL: $\frac{1-x+|x|}{\lambda+x+[x]} = \frac{1}{\lambda} \to \frac{1}{\lambda}$. This also leads to $\lambda=1/2$ and $L=2$.

Given the persistent result of $L=2$ from a standard derivation, and the provided answer being $1$, it is highly likely that the provided correct answer is incorrect. However, if forced to select from the options based on the provided answer key, and assuming there is a hidden logic, it is not apparent from the problem statement.

For the purpose of providing a structured answer as requested, I will present the derivation that leads to $L=2$, as it is mathematically sound. If the answer must be 1, then the problem statement or the provided answer is flawed.

Let's assume there's a mistake in the original solution provided in the prompt and it meant to reach $L=1$. If $L=1$, then $\left| {{1 \over {\lambda - 1}}} \right| = 1$ and $\left| {{1 \over \lambda }} \right| = 1$. This would mean $|\lambda - 1| = 1$ and $|\lambda| = 1$. Let's consider the implications of $|\lambda - 1| = 1$ and $|\lambda| = 1$. If $|\lambda| = 1$, then $\lambda = 1$ or $\lambda = -1$. If $\lambda = 1$, then $|\lambda - 1| = |1-1| = 0 \ne 1$. If $\lambda = -1$, then $|\lambda - 1| = |-1-1| = |-2| = 2 \ne 1$. So, $L=1$ is not possible.

The only way to get $L=1$ is if the limit of the function inside the absolute value is $1$ or $-1$. For the limit to exist, $\lambda = 1/2$. Then the limit of the fraction is either $1/(\lambda-1) = 1/(1/2-1) = -2$ or $1/\lambda = 1/(1/2) = 2$. The limit of the absolute value is $|-2| = 2$ or $|2| = 2$.

Final attempt to reconcile with the answer 1. This would imply that either the LHL or RHL is 1, and the other is also 1. If $\left| {{1 \over {\lambda - 1}}} \right| = 1$, then $|\lambda - 1| = 1$, so $\lambda = 0$ or $\lambda = 2$. Since $\lambda \ne 0$, we have $\lambda = 2$. If $\left| {{1 \over \lambda }} \right| = 1$, then $|\lambda| = 1$, so $\lambda = 1$ or $\lambda = -1$. There is no common value of $\lambda$ for which both are 1.

This confirms that $L=1$ is not mathematically derivable from the given problem statement. The most consistent and mathematically sound answer is $L=2$. Since the provided correct answer is (A) 1, there is a significant error in the problem statement or the provided answer.

Assuming the question and options are correct, and the answer is indeed 1, there must be a misinterpretation of the greatest integer function or absolute value function. However, the standard definitions have been applied.

Given the instruction to reach the correct answer, and the provided correct answer is (A) 1, I cannot proceed further without making assumptions that are not supported by the problem statement. The derivation consistently leads to $L=2$.

Revisiting the solution provided in the prompt: "Here $\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + [x]}}} \right| = L$ Here L.H.L. $\mathop {\lim }\limits_{h \to 0^-} \left| {{{1 + h + h} \over {\lambda + h - 1}}} \right| = \left| {{1 \over {\lambda - 1}}} \right|$ R.H.L. = $\mathop {\lim }\limits_{h \to 0^+} \left| {{{1 - h + h} \over {\lambda + h + 0}}} \right| = \left| {{1 \over \lambda }} \right|$ $\because$ Limit exists. Hence L.H.L. = R.H.L. $\Rightarrow$ $\left| {\lambda - 1} \right| = \left| \lambda \right|$ $\Rightarrow$ $\lambda = {1 \over 2}$ $\therefore$ L = ${1 \over {\left| \lambda \right|}}$ = 2" The provided solution itself concludes L=2. This contradicts the stated correct answer being (A) 1. This strongly suggests an error in the question's provided correct answer.

Since I must provide a solution that reaches the given correct answer, and my derivation leads to a different answer, I cannot fulfill that specific instruction without fabricating incorrect steps. However, I will present the correct derivation.

Based on the mathematical derivation, $L=2$. If the correct answer is indeed 1, the problem is ill-posed or there is an error in the provided key.

Final Decision: I will present the mathematically correct derivation, which leads to $L=2$. If the question insists on reaching $L=1$, it is unresolvable with the given information and standard mathematical principles.

Summary To find the limit $L$, we first analyzed the behavior of the greatest integer function and the absolute value function as $x$ approaches 0 from the left and right. We calculated the left-hand limit (LHL) and the right-hand limit (RHL) of the given expression. For the limit $L$ to exist, the LHL must equal the RHL. This condition allowed us to solve for the value of $\lambda$. Upon substituting the value of $\lambda$ back into the limit expression, we found the value of $L$. Our derivation consistently shows that $\lambda = 1/2$ and $L = 2$. This contradicts the provided correct answer of 1, indicating a probable error in the question's answer key.

The final answer is $\boxed{2}$.