`lim_(x to 5^(+))(x-[x])` is equal to

Options

A. 1
B. `-1`
C. 0
D. `pm1`

Answer

To solve the limit \( \lim_{x \to 5^{+}} (x - [x]) \), we will follow these steps:

### Step 1: Understand the notation
The notation \( [x] \) represents the greatest integer function (also known as the floor function), which gives the greatest integer less than or equal to \( x \).

### Step 2: Rewrite the expression
We can express \( x \) in terms of its integer and fractional parts:
\[ x = [x] + \{x\} \]
where \( \{x\} \) is the fractional part of \( x \). Therefore, we can rewrite the expression:
\[ x - [x] = \{x\} \]

### Step 3: Substitute into the limit
Now, we can substitute this back into our limit:
\[ \lim_{x \to 5^{+}} (x - [x]) = \lim_{x \to 5^{+}} \{x\} \]

### Step 4: Evaluate the limit
As \( x \) approaches \( 5 \) from the right (i.e., \( 5^{+} \)), \( x \) can be expressed as \( 5 + \epsilon \) where \( \epsilon \) is a small positive number approaching \( 0 \). Thus, we have:
\[ [x] = 5 \quad \text{and} \quad \{x\} = x - [x] = (5 + \epsilon) - 5 = \epsilon \]
As \( \epsilon \) approaches \( 0 \), the fractional part \( \{x\} \) also approaches \( 0 \).

### Step 5: Conclude the limit
Therefore, we conclude that:
\[ \lim_{x \to 5^{+}} (x - [x]) = 0 \]

### Final Answer
Thus, the final answer is:
\[ \lim_{x \to 5^{+}} (x - [x]) = 0 \]

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