Evaluate `lim_(x to 3) (sqrt(x + 3) - sqrt(6))/(x^(2) - 9)`

Answer

To evaluate the limit

\[ \lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}, \]

we will follow these steps:

### Step 1: Substitute the limit value
First, we substitute \( x = 3 \) into the expression:

\[ \text{Numerator: } \sqrt{3 + 3} - \sqrt{6} = \sqrt{6} - \sqrt{6} = 0, \]
\[ \text{Denominator: } 3^2 - 9 = 9 - 9 = 0. \]

Both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \).

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

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

if the limit on the right exists. We will differentiate the numerator and the denominator.

**Hint:** Differentiate the numerator and denominator separately.

### Step 3: Differentiate the numerator and denominator
1. **Differentiate the numerator:**
\[ \frac{d}{dx}(\sqrt{x + 3} - \sqrt{6}) = \frac{1}{2\sqrt{x + 3}} - 0 = \frac{1}{2\sqrt{x + 3}}. \]

2. **Differentiate the denominator:**
\[ \frac{d}{dx}(x^2 - 9) = 2x. \]

### Step 4: Rewrite the limit using derivatives
Now we rewrite the limit using the derivatives:

\[ \lim_{x \to 3} \frac{\frac{1}{2\sqrt{x + 3}}}{2x}. \]

**Hint:** Substitute \( x = 3 \) again after differentiating.

### Step 5: Substitute \( x = 3 \) into the new limit
Substituting \( x = 3 \):

\[ = \frac{\frac{1}{2\sqrt{3 + 3}}}{2 \cdot 3} = \frac{\frac{1}{2\sqrt{6}}}{6} = \frac{1}{2\sqrt{6} \cdot 6} = \frac{1}{12\sqrt{6}}. \]

### Final Result
Thus, the limit evaluates to:

\[ \lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9} = \frac{1}{12\sqrt{6}}. \]