`lim_(x to 3)(x-3)/(sqrt(x-2)-sqrt(4-x))`=

Options

A. 2
B. 3
C. 1
D. 0

Answer

To solve the limit \( \lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}} \), we can follow these steps:

### Step 1: Substitute the limit value
First, we substitute \( x = 3 \) into the expression to check if it results in an indeterminate form:
\[ \frac{3 - 3}{\sqrt{3 - 2} - \sqrt{4 - 3}} = \frac{0}{1 - 1} = \frac{0}{0} \]
Since we have an indeterminate form \( \frac{0}{0} \), we need to manipulate the expression.

### Step 2: Rationalize the denominator
To eliminate the square roots in the denominator, we can multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}} \cdot \frac{\sqrt{x - 2} + \sqrt{4 - x}}{\sqrt{x - 2} + \sqrt{4 - x}} = \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{(\sqrt{x - 2})^2 - (\sqrt{4 - x})^2} \]

### Step 3: Simplify the denominator
Now, we simplify the denominator using the difference of squares:
\[ (\sqrt{x - 2})^2 - (\sqrt{4 - x})^2 = (x - 2) - (4 - x) = x - 2 - 4 + x = 2x - 6 \]
Thus, we have:
\[ \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2x - 6} \]

### Step 4: Factor the denominator
Notice that \( 2x - 6 \) can be factored:
\[ 2x - 6 = 2(x - 3) \]
So, we can rewrite our limit as:
\[ \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2(x - 3)} \]

### Step 5: Cancel out the common terms
Now, we can cancel \( x - 3 \) from the numerator and the denominator:
\[ \frac{\sqrt{x - 2} + \sqrt{4 - x}}{2} \]

### Step 6: Substitute the limit value again
Now we can substitute \( x = 3 \) into the simplified expression:
\[ \frac{\sqrt{3 - 2} + \sqrt{4 - 3}}{2} = \frac{\sqrt{1} + \sqrt{1}}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}} = 1 \]