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The given problem asks to find the limit:

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

This is a typical limit problem involving a difference of square roots. Let's go step by step to solve it:

### Step 1: Check direct substitution
First, substitute \( x = 3 \) directly into the expression:

\[
\frac{\sqrt{6(3) - 2} - \sqrt{3(3) + 7}}{3 - 3}
= \frac{\sqrt{18 - 2} - \sqrt{9 + 7}}{0}
= \frac{\sqrt{16} - \sqrt{16}}{0}
= \frac{4 - 4}{0}
= \frac{0}{0}
\]

This results in an indeterminate form \( \frac{0}{0} \), so we need to simplify using other techniques, such as multiplying by the conjugate.

### Step 2: Multiply by the conjugate
To resolve the indeterminate form, multiply both the numerator and the denominator by the conjugate of the numerator:

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

This will simplify the numerator using the difference of squares:

\[
= \frac{(6x - 2) - (3x + 7)}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})}
= \frac{6x - 2 - 3x - 7}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})}
= \frac{3x - 9}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})}
\]

### Step 3: Factor and cancel terms
The numerator \( 3x - 9 \) can be factored:

\[
= \frac{3(x - 3)}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})}
\]

Now, cancel the \( x - 3 \) term in both the numerator and denominator:

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

### Step 4: Substitute \( x = 3 \) into the simplified expression
Now, substitute \( x = 3 \) into the simplified expression:

\[
= \frac{3}{\sqrt{6(3) - 2} + \sqrt{3(3) + 7}}
= \frac{3}{\sqrt{18 - 2} + \sqrt{9 + 7}}
= \frac{3}{\sqrt{16} + \sqrt{16}}
= \frac{3}{4 + 4}
= \frac{3}{8}
\]

Thus, the value of the limit is:

\[
\boxed{\frac{3}{8}}
\]

The correct answer is **C**.

---

Do you have any further questions, or would you like more details on any step? Here are some related questions you could explore:

1. How can limits involving square roots be simplified in general?
2. What is the difference of squares and how is it used in limit problems?
3. What happens if a limit does not yield an indeterminate form like \( \frac{0}{0} \)?
4. Can L'Hôpital's Rule be applied in this case? Why or why not?
5. How do you handle limits at infinity involving square roots?

**Tip:** Always check for indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying simplification techniques like conjugates or factoring.

Nilai lim x->3 √(6x-2)-√(3x+7) / (x-3) = ....

Learn how to solve the limit lim x->3 √(6x-2) - √(3x+7) / (x-3) using conjugates to simplify square roots and resolve indeterminate forms. This detailed step-by-step solution explores advanced calculus concepts such as limits, square roots, and the difference of squares theorem. Suitable for students in Grades 11-12.

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