If [x] denotes the greatest integer less than or equal to x,then the value of `lim_(x->1)(1-x+[x-1]+[1-x])` is

Options

A. 0
B. 1
C. `-1`
D. none of these

Answer

To solve the limit \( \lim_{x \to 1} (1 - x + [x - 1] + [1 - x]) \), where \([x]\) denotes the greatest integer less than or equal to \(x\), we will calculate the right-hand limit (RHL) and the left-hand limit (LHL) separately.

### Step 1: Calculate the Right-Hand Limit (RHL)

We start by calculating the right-hand limit as \(x\) approaches 1 from the right (\(x \to 1^+\)):

\[ \lim_{x \to 1^+} (1 - x + [x - 1] + [1 - x]) \]

1. **Evaluate \(1 - x\)**:
- As \(x\) approaches 1 from the right, \(1 - x\) approaches \(0\).

2. **Evaluate \([x - 1]\)**:
- For \(x \to 1^+\), \(x - 1\) approaches \(0^+\) (a small positive number). Therefore, \([x - 1] = 0\).

3. **Evaluate \([1 - x]\)**:
- As \(x\) approaches 1 from the right, \(1 - x\) approaches \(0^-\) (a small negative number). Thus, \([1 - x] = -1\).

Putting these values together:

\[ \lim_{x \to 1^+} (1 - x + [x - 1] + [1 - x]) = 0 + 0 - 1 = -1 \]

### Step 2: Calculate the Left-Hand Limit (LHL)

Next, we calculate the left-hand limit as \(x\) approaches 1 from the left (\(x \to 1^-\)):

\[ \lim_{x \to 1^-} (1 - x + [x - 1] + [1 - x]) \]

1. **Evaluate \(1 - x\)**:
- As \(x\) approaches 1 from the left, \(1 - x\) approaches \(0\).

2. **Evaluate \([x - 1]\)**:
- For \(x \to 1^-\), \(x - 1\) approaches \(0^-\) (a small negative number). Therefore, \([x - 1] = -1\).

3. **Evaluate \([1 - x]\)**:
- As \(x\) approaches 1 from the left, \(1 - x\) approaches \(0^+\) (a small positive number). Thus, \([1 - x] = 0\).

Putting these values together:

\[ \lim_{x \to 1^-} (1 - x + [x - 1] + [1 - x]) = 0 - 1 + 0 = -1 \]

### Step 3: Conclusion

Since both the right-hand limit and left-hand limit are equal:

\[ \lim_{x \to 1^+} = -1 \quad \text{and} \quad \lim_{x \to 1^-} = -1 \]

We conclude:

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

Thus, the final answer is:

\[ \boxed{-1} \]