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To solve the given limit problem:

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

### Step 1: Direct substitution
If we directly substitute \( x = 2 \) into the expression, we get:

\[
\frac{\sqrt{3(2)} - \sqrt{6}}{2 - 2} = \frac{\sqrt{6} - \sqrt{6}}{0} = \frac{0}{0}
\]

This results in an indeterminate form, \( \frac{0}{0} \), so we need to use another method to evaluate the limit. One common method for this kind of limit is **rationalizing the numerator**.

### Step 2: Rationalize the numerator
We multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{3x} + \sqrt{6} \). This gives:

\[
\lim_{{x \to 2}} \frac{\sqrt{3x} - \sqrt{6}}{x - 2} \cdot \frac{\sqrt{3x} + \sqrt{6}}{\sqrt{3x} + \sqrt{6}} = \lim_{{x \to 2}} \frac{(\sqrt{3x})^2 - (\sqrt{6})^2}{(x - 2)(\sqrt{3x} + \sqrt{6})}
\]

### Step 3: Simplify the expression
Now simplify the numerator:

\[
(\sqrt{3x})^2 = 3x \quad \text{and} \quad (\sqrt{6})^2 = 6
\]

Thus, the numerator becomes:

\[
3x - 6
\]

So the expression is now:

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

Factor the numerator:

\[
3x - 6 = 3(x - 2)
\]

Now the expression becomes:

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

Cancel out the \( x - 2 \) terms:

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

### Step 4: Substitute \( x = 2 \) into the simplified expression
Now substitute \( x = 2 \):

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

Finally, simplify the expression:

\[
\frac{3}{2\sqrt{6}} = \frac{3}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{6}}{12} = \frac{\sqrt{6}}{4}
\]

### Final Answer:
The value of the limit is:

\[
\frac{\sqrt{6}}{4}
\]

Would you like further details or have any questions?

Here are 5 questions related to the topic:
1. How does rationalizing the numerator help in solving limits involving radicals?
2. What other techniques can be used to solve indeterminate limits?
3. Can L'Hôpital's Rule be applied to this problem?
4. How do you simplify radical expressions in limits?
5. Why is the direct substitution method insufficient for some limits?

**Tip:** Rationalizing expressions is particularly helpful when dealing with square roots in limits that result in indeterminate forms like \( \frac{0}{0} \).

lim(x → 2) [sqrt(3x) - sqrt(6)] / (x - 2)

This guide provides a detailed solution to the limit problem lim(x → 2) [sqrt(3x) - sqrt(6)] / (x - 2). Learn how to use rationalization and other algebraic techniques to solve indeterminate forms like 0/0. Suitable for Grades 11-12 students.

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