`lim_(x to 5^(+)) (x^(2)-9x +20)/(x-[x])` is equal to : (`[*]` represent greatest integer function)

Options

A. 0
B. 1
C. 2
D. Not define

Answer

To solve the limit \( \lim_{x \to 5^{+}} \frac{x^{2} - 9x + 20}{x - [x]} \), where \([x]\) represents the greatest integer function, we will go through the following steps:

### Step 1: Understand the behavior of the greatest integer function
As \( x \) approaches \( 5^{+} \), \( x \) is slightly greater than \( 5 \) (for example, \( 5.0001 \)). The greatest integer function \([x]\) will give us \( 5 \) since it returns the largest integer less than or equal to \( x \).

**Hint:** When \( x \) is slightly greater than \( 5 \), \([x] = 5\).

### Step 2: Substitute the value of \([x]\) in the limit
Now we can rewrite the limit:
\[ \lim_{x \to 5^{+}} \frac{x^{2} - 9x + 20}{x - 5} \]

**Hint:** Replace \([x]\) with \( 5 \) in the limit expression.

### Step 3: Factor the numerator
Next, we need to factor the quadratic expression in the numerator:
\[ x^{2} - 9x + 20 = (x - 4)(x - 5) \]

**Hint:** Look for two numbers that multiply to \( 20 \) and add to \( -9 \).

### Step 4: Rewrite the limit with the factored form
Now we can rewrite the limit as:
\[ \lim_{x \to 5^{+}} \frac{(x - 4)(x - 5)}{x - 5} \]

**Hint:** Notice that \( (x - 5) \) appears in both the numerator and the denominator.

### Step 5: Cancel the common term
Since \( x \to 5^{+} \) means \( x \) is not equal to \( 5 \), we can safely cancel \( (x - 5) \):
\[ \lim_{x \to 5^{+}} (x - 4) \]

**Hint:** Ensure that \( x \) is approaching but not equal to \( 5 \) to avoid division by zero.

### Step 6: Evaluate the limit
Now we can directly substitute \( x = 5 \):
\[ 5 - 4 = 1 \]

**Hint:** After simplifying, directly substitute the limit value into the remaining expression.

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 5^{+}} \frac{x^{2} - 9x + 20}{x - [x]} = 1 \]