`lim_(xrarr2) (sqrt(3-x)-1)/(2-x)`

Answer

To solve the limit problem \( \lim_{x \to 2} \frac{\sqrt{3-x} - 1}{2-x} \), we will follow these steps:

### Step 1: Substitute the limit value
First, we substitute \( x = 2 \) into the expression:

\[ \frac{\sqrt{3-2} - 1}{2-2} = \frac{\sqrt{1} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]

This is an indeterminate form \( \frac{0}{0} \), which means we need to apply L'Hôpital's Rule.

**Hint:** When you encounter \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), consider using L'Hôpital's Rule.

### Step 2: Differentiate the numerator and denominator
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

- The numerator is \( \sqrt{3-x} - 1 \). The derivative is:
\[ \frac{d}{dx}(\sqrt{3-x}) = \frac{1}{2\sqrt{3-x}} \cdot (-1) = -\frac{1}{2\sqrt{3-x}} \]

- The denominator is \( 2-x \). The derivative is:
\[ \frac{d}{dx}(2-x) = -1 \]

Now we can rewrite the limit using these derivatives:

\[ \lim_{x \to 2} \frac{-\frac{1}{2\sqrt{3-x}}}{-1} = \lim_{x \to 2} \frac{1}{2\sqrt{3-x}} \]

**Hint:** Remember to differentiate both the numerator and denominator when applying L'Hôpital's Rule.

### Step 3: Substitute the limit value again
Now, we substitute \( x = 2 \) into the new limit expression:

\[ \lim_{x \to 2} \frac{1}{2\sqrt{3-x}} = \frac{1}{2\sqrt{3-2}} = \frac{1}{2\sqrt{1}} = \frac{1}{2 \cdot 1} = \frac{1}{2} \]

**Hint:** After differentiating, always substitute the limit value again to find the final answer.

### Final Answer
Thus, the limit is:

\[ \boxed{1} \]

**Note:** The final answer mentioned in the video transcript is incorrect; the correct limit is \( \frac{1}{2} \).