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The given problem is asking for the limit:

\[
\lim_{h \to 0} \frac{\sqrt{3(a + h)} - \sqrt{3a}}{h}
\]

### Solution Outline:

This is a standard limit that can be solved using the method of rationalization. Here's how we approach it:

1. **Multiply by the conjugate**: 
   Multiply both the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{3(a + h)} + \sqrt{3a}\), to eliminate the square roots.

\[
\frac{\sqrt{3(a + h)} - \sqrt{3a}}{h} \cdot \frac{\sqrt{3(a + h)} + \sqrt{3a}}{\sqrt{3(a + h)} + \sqrt{3a}}
\]

2. **Simplify the expression**:
   The numerator will become:

\[
\left(\sqrt{3(a + h)}\right)^2 - \left(\sqrt{3a}\right)^2 = 3(a + h) - 3a = 3h
\]

Now, the expression becomes:

\[
\frac{3h}{h \left(\sqrt{3(a + h)} + \sqrt{3a}\right)}
\]

3. **Cancel out the \(h\) terms**:
   The \(h\) in the numerator and denominator cancels out, leaving:

\[
\frac{3}{\sqrt{3(a + h)} + \sqrt{3a}}
\]

4. **Take the limit as \(h \to 0\)**:
   As \(h\) approaches 0, \(\sqrt{3(a + h)}\) approaches \(\sqrt{3a}\), so the expression simplifies to:

\[
\frac{3}{2\sqrt{3a}}
\]

Thus, the limit is:

\[
\frac{1}{2\sqrt{a}}
\]

### Final Answer:

The limit is:

\[
\boxed{\frac{1}{2\sqrt{a}}}
\]

Let me know if you'd like further clarification or additional steps explained! Here are 5 related questions for deeper understanding:

1. How would the approach change if the limit were taken as \(h \to \infty\)?
2. How does rationalization work in solving limits with square roots?
3. Can we apply L'Hopital's Rule to solve this limit? How?
4. What are other techniques to solve indeterminate forms in limits?
5. How would the result change if the square roots involved different constants?

**Tip**: Whenever you encounter square roots in a limit, multiplying by the conjugate often simplifies the expression and helps solve the problem.

Find (in terms of the constant a) \[ \lim_ {h \to 0} \frac{ \sqrt{3(a + h)} - \sqrt{3a}}{h} \].

Explore the limit \( \lim_ {h \to 0} \frac{ \sqrt{3(a + h)} - \sqrt{3a}}{h} \) using rationalization to simplify the expression. This example covers key calculus concepts and the method of rationalization, suitable for high school calculus students.

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