If `lim_(xto1) (2-x+a[x-1]+b[1+x])` exists, then a and b can take the values (where `[.]` denotes the greatest integer function)

Explanation

To solve the problem, we need to determine the values of \( a \) and \( b \) such that the limit

\[ \lim_{x \to 1} (2 - x + a[\lfloor x - 1 \rfloor] + b[\lfloor x + 1 \rfloor]) \]

exists. Here, \( \lfloor . \rfloor \) denotes the greatest integer function.

### Step-by-Step Solution

**Step 1: Find the Left-Hand Limit**

We start by calculating the left-hand limit as \( x \) approaches 1 from the left, denoted as \( x \to 1^- \).

\[ \lim_{x \to 1^-} (2 - x + a[\lfloor x - 1 \rfloor] + b[\lfloor x + 1 \rfloor]) \]

For \( x \) slightly less than 1 (i.e., \( x = 1 - \epsilon \) where \( \epsilon \) is a small positive number):

- \( \lfloor x - 1 \rfloor = \lfloor 1 - \epsilon - 1 \rfloor = \lfloor -\epsilon \rfloor = -1 \) (since \( -\epsilon < 0 \))
- \( \lfloor x + 1 \rfloor = \lfloor 1 - \epsilon + 1 \rfloor = \lfloor 2 - \epsilon \rfloor = 1 \) (since \( 2 - \epsilon < 2 \))

Substituting these values into the limit expression:

\[ \lim_{x \to 1^-} (2 - (1 - \epsilon) + a(-1) + b(1)) \]

This simplifies to:

\[ \lim_{x \to 1^-} (1 + \epsilon - a + b) = 1 - a + b \]

**Step 2: Find the Right-Hand Limit**

Next, we calculate the right-hand limit as \( x \) approaches 1 from the right, denoted as \( x \to 1^+ \).

\[ \lim_{x \to 1^+} (2 - x + a[\lfloor x - 1 \rfloor] + b[\lfloor x + 1 \rfloor]) \]

For \( x \) slightly more than 1 (i.e., \( x = 1 + \epsilon \)):

- \( \lfloor x - 1 \rfloor = \lfloor 1 + \epsilon - 1 \rfloor = \lfloor \epsilon \rfloor = 0 \) (since \( \epsilon > 0 \))
- \( \lfloor x + 1 \rfloor = \lfloor 1 + \epsilon + 1 \rfloor = \lfloor 2 + \epsilon \rfloor = 2 \) (since \( 2 + \epsilon > 2 \))

Substituting these values into the limit expression:

\[ \lim_{x \to 1^+} (2 - (1 + \epsilon) + a(0) + b(2)) \]

This simplifies to:

\[ \lim_{x \to 1^+} (1 - \epsilon + 2b) = 1 + 2b \]

**Step 3: Set Left-Hand Limit Equal to Right-Hand Limit**

For the limit to exist, the left-hand limit must equal the right-hand limit:

\[ 1 - a + b = 1 + 2b \]

Subtracting 1 from both sides gives:

\[ -a + b = 2b \]

Rearranging this gives:

\[ -a = b \]

Thus, we find the relationship:

\[ b = -a \]

### Conclusion

The values of \( a \) and \( b \) can take any values as long as they satisfy the relationship \( b = -a \).