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underset(x to 2)(Lt) {[x - 2] + [2 - x] ...

`underset(x to 2)(Lt) {[x - 2] + [2 - x] - x} =` where [.] represents greater intergral function

A

0

B

3

C

`-3`

D

`-2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the limit problem \(\lim_{x \to 2} \left[ \lfloor x - 2 \rfloor + \lfloor 2 - x \rfloor - x \right]\), where \(\lfloor . \rfloor\) represents the greatest integer function, we will evaluate the limit from both the right-hand side (as \(x\) approaches 2 from values greater than 2) and the left-hand side (as \(x\) approaches 2 from values less than 2). ### Step 1: Evaluate the Right-Hand Limit (\(x \to 2^+\)) 1. **Substituting \(x = 2^+\)**: - \(x - 2\) approaches \(0^+\) (a small positive number). - \(2 - x\) approaches \(0^-\) (a small negative number). 2. **Calculate \(\lfloor x - 2 \rfloor\)**: - Since \(x - 2\) is a small positive number, \(\lfloor x - 2 \rfloor = 0\). 3. **Calculate \(\lfloor 2 - x \rfloor\)**: - Since \(2 - x\) is a small negative number, \(\lfloor 2 - x \rfloor = -1\). 4. **Substituting into the limit expression**: \[ \lfloor x - 2 \rfloor + \lfloor 2 - x \rfloor - x = 0 - 1 - 2 = -3 \] ### Step 2: Evaluate the Left-Hand Limit (\(x \to 2^-\)) 1. **Substituting \(x = 2^-\)**: - \(x - 2\) approaches \(0^-\) (a small negative number). - \(2 - x\) approaches \(0^+\) (a small positive number). 2. **Calculate \(\lfloor x - 2 \rfloor\)**: - Since \(x - 2\) is a small negative number, \(\lfloor x - 2 \rfloor = -1\). 3. **Calculate \(\lfloor 2 - x \rfloor\)**: - Since \(2 - x\) is a small positive number, \(\lfloor 2 - x \rfloor = 0\). 4. **Substituting into the limit expression**: \[ \lfloor x - 2 \rfloor + \lfloor 2 - x \rfloor - x = -1 + 0 - 2 = -3 \] ### Step 3: Conclusion Both the right-hand limit and the left-hand limit yield the same value: \[ \lim_{x \to 2^+} \left[ \lfloor x - 2 \rfloor + \lfloor 2 - x \rfloor - x \right] = -3 \] \[ \lim_{x \to 2^-} \left[ \lfloor x - 2 \rfloor + \lfloor 2 - x \rfloor - x \right] = -3 \] Thus, the overall limit is: \[ \lim_{x \to 2} \left[ \lfloor x - 2 \rfloor + \lfloor 2 - x \rfloor - x \right] = -3 \] ### Final Answer: \(-3\)
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