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The problem is asking to find the limit of the function:

\[
\lim_{x \to 3} \frac{\sqrt{6x - 2} - \sqrt{3x + 7}}{x - 3}
\]

This is an indeterminate form of \(\frac{0}{0}\) when directly substituting \(x = 3\), so we will need to apply algebraic techniques, such as multiplying by the conjugate, to simplify the expression and resolve the limit.

### Step-by-Step Solution:

1. **Rewrite the expression:**
   \[
   \lim_{x \to 3} \frac{\sqrt{6x - 2} - \sqrt{3x + 7}}{x - 3}
   \]

2. **Multiply numerator and denominator by the conjugate:**
   Multiply both the numerator and the denominator by the conjugate of the numerator, which is \(\sqrt{6x - 2} + \sqrt{3x + 7}\):
   \[
   \lim_{x \to 3} \frac{\left(\sqrt{6x - 2} - \sqrt{3x + 7}\right) \cdot \left(\sqrt{6x - 2} + \sqrt{3x + 7}\right)}{\left(x - 3\right) \cdot \left(\sqrt{6x - 2} + \sqrt{3x + 7}\right)}
   \]

3. **Simplify the numerator:**
   Using the difference of squares formula, the numerator becomes:
   \[
   \left(\sqrt{6x - 2}\right)^2 - \left(\sqrt{3x + 7}\right)^2 = (6x - 2) - (3x + 7) = 6x - 2 - 3x - 7 = 3x - 9
   \]
   Therefore, the limit simplifies to:
   \[
   \lim_{x \to 3} \frac{3x - 9}{(x - 3) \cdot \left(\sqrt{6x - 2} + \sqrt{3x + 7}\right)}
   \]

4. **Factor the numerator:**
   Factor \(3x - 9\) as:
   \[
   3(x - 3)
   \]
   So the limit becomes:
   \[
   \lim_{x \to 3} \frac{3(x - 3)}{(x - 3) \cdot \left(\sqrt{6x - 2} + \sqrt{3x + 7}\right)}
   \]

5. **Cancel the \((x - 3)\) terms:**
   Since \((x - 3)\) appears in both the numerator and the denominator, we can cancel these terms:
   \[
   \lim_{x \to 3} \frac{3}{\sqrt{6x - 2} + \sqrt{3x + 7}}
   \]

6. **Substitute \(x = 3\):**
   Now substitute \(x = 3\) into the simplified expression:
   \[
   \frac{3}{\sqrt{6(3) - 2} + \sqrt{3(3) + 7}} = \frac{3}{\sqrt{18 - 2} + \sqrt{9 + 7}} = \frac{3}{\sqrt{16} + \sqrt{16}} = \frac{3}{4 + 4} = \frac{3}{8}
   \]

### Final Answer:

The value of the limit is:
\[
\boxed{\frac{3}{8}}
\]

Do you have any questions or want more details on any step?

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Here are 5 related questions to enhance understanding:

1. What is the difference between an indeterminate form and a determinate form?
2. How does multiplying by the conjugate help in solving limit problems?
3. Can this limit be solved using L'Hopital's Rule instead? How?
4. What other methods can be used to solve limits involving square roots?
5. How would the limit change if the expression was slightly different, such as \(\sqrt{6x + 2} - \sqrt{3x - 7}\)?

**Tip:** Always check for indeterminate forms when calculating limits, and apply the appropriate technique like factoring, multiplying by the conjugate, or L'Hopital's Rule.

Find the value of the limit: lim (x -> 3) (√(6x - 2) - √(3x + 7)) / (x - 3).

Learn how to solve the limit of a square root function using the conjugate method. This step-by-step solution helps you evaluate lim (x -> 3) (√(6x - 2) - √(3x + 7)) / (x - 3). Suitable for students in Grades 10-12, this problem covers limit laws, indeterminate forms, and algebraic techniques.

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