`lim_(xrarr2) (x^(2)-4)/(sqrt(x+2)-sqrt(3x-2))`

Answer

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

### Step 1: Substitute \( x = 2 \)
First, we substitute \( x = 2 \) into the expression to check if it results in an indeterminate form.

\[ \text{Numerator: } x^2 - 4 = 2^2 - 4 = 4 - 4 = 0 \]
\[ \text{Denominator: } \sqrt{x + 2} - \sqrt{3x - 2} = \sqrt{2 + 2} - \sqrt{3(2) - 2} = \sqrt{4} - \sqrt{6 - 2} = 2 - 2 = 0 \]

Since both the numerator and denominator evaluate to 0, we have the indeterminate form \( \frac{0}{0} \).

**Hint:** Check if substituting the limit value gives an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

**Numerator:**
\[ \frac{d}{dx}(x^2 - 4) = 2x \]

**Denominator:**
To differentiate the denominator \( \sqrt{x + 2} - \sqrt{3x - 2} \), we use the chain rule:
\[ \frac{d}{dx}(\sqrt{x + 2}) = \frac{1}{2\sqrt{x + 2}} \cdot 1 = \frac{1}{2\sqrt{x + 2}} \]
\[ \frac{d}{dx}(\sqrt{3x - 2}) = \frac{1}{2\sqrt{3x - 2}} \cdot 3 = \frac{3}{2\sqrt{3x - 2}} \]
Thus, the derivative of the denominator is:
\[ \frac{1}{2\sqrt{x + 2}} - \frac{3}{2\sqrt{3x - 2}} \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 2} \frac{2x}{\frac{1}{2\sqrt{x + 2}} - \frac{3}{2\sqrt{3x - 2}}} \]

### Step 4: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \) into the new expression:
\[ \text{Numerator: } 2(2) = 4 \]
\[ \text{Denominator: } \frac{1}{2\sqrt{2 + 2}} - \frac{3}{2\sqrt{3(2) - 2}} = \frac{1}{2\sqrt{4}} - \frac{3}{2\sqrt{4}} = \frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2} \]

### Step 5: Calculate the limit
Now we can calculate the limit:
\[ \lim_{x \to 2} \frac{4}{-\frac{1}{2}} = 4 \cdot -2 = -8 \]

### Final Answer
Thus, the limit is:
\[ \boxed{-8} \]