x→1 lim ​ (1−x+[x−1]+[1−x]) is equal to (where [.] denotes greatest integer function)

Explanation

To solve the limit problem

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

where \([.]\) denotes the greatest integer function, we will evaluate the left-hand limit (LHL) and the right-hand limit (RHL) separately.

### Step 1: Evaluate the Left-Hand Limit (LHL)

As \(x\) approaches 1 from the left (\(x \to 1^{-}\)), we have:

- \(1 - x\) approaches \(0\).
- \(x - 1\) approaches \(0\) from the negative side, so \([x - 1] = -1\) (since the greatest integer less than or equal to a negative number just below zero is -1).
- \(1 - x\) approaches \(0\) from the negative side, so \([1 - x] = 0\) (since the greatest integer less than or equal to a negative number just below zero is 0).

Now substituting these values into the limit expression:

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

### Step 2: Evaluate the Right-Hand Limit (RHL)

As \(x\) approaches 1 from the right (\(x \to 1^{+}\)), we have:

- \(1 - x\) approaches \(0\).
- \(x - 1\) approaches \(0\) from the positive side, so \([x - 1] = 0\) (since the greatest integer less than or equal to a positive number just above zero is 0).
- \(1 - x\) approaches \(0\) from the negative side, so \([1 - x] = -1\).

Now substituting these values into the limit expression:

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

### Step 3: Conclusion

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

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

Thus, the final answer is:

\[ \boxed{-1} \]