To solve the limit

\[
\lim_{x \to 0} \frac{\sqrt{9 + x} - 3}{2 + 2x},
\]

we notice that direct substitution of \( x = 0 \) into the expression results in the indeterminate form \( \frac{0}{2} \), which is not indeterminate in the usual sense but indicates a need for further simplification.

### Step 1: Simplify the Expression
Start by simplifying the numerator. The numerator \( \sqrt{9 + x} - 3 \) suggests we could use the technique of rationalization. Multiply both the numerator and denominator by the conjugate of the numerator:

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

Simplify the numerator:

\[
\frac{x}{(2 + 2x)(\sqrt{9 + x} + 3)}.
\]

### Step 2: Take the Limit
Now, as \( x \) approaches 0, substitute \( x = 0 \) into the expression:

\[
\lim_{x \to 0} \frac{x}{(2 + 2x)(\sqrt{9 + x} + 3)} = \frac{0}{(2 + 0)(\sqrt{9 + 0} + 3)} = \frac{0}{2 \times 6} = \frac{0}{12} = 0.
\]

Thus, the limit is \(\boxed{0}\).

Do you want details or have any questions? Here are some related questions to explore:

1. Why is rationalization useful in solving this limit problem?
2. What happens if you use L'Hôpital's Rule on this limit?
3. Can you find another limit where rationalization simplifies the solution?
4. How does the presence of \( 2x \) in the denominator affect the limit?
5. What would the limit be if the denominator was just a constant, like \( 2 \)?

**Tip:** Rationalizing the numerator or denominator is a key technique when dealing with square roots in limits, especially when you encounter indeterminate forms.

Learn how to solve the limit problem lim x → 0 (sqrt(9 + x) - 3) / (2 + 2x) with a step-by-step solution. This problem involves understanding limits and the technique of rationalization in calculus. Suitable for advanced high school students and college-level calculus enthusiasts.

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