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To solve the limit:

\[
\lim_{{x \to 4}} \frac{\sqrt{3x + 6}}{-3x}
\]

we can proceed with direct substitution to see if it gives us a result, or if further simplification is needed.

1. **Direct Substitution:**
   Substitute \( x = 4 \) directly into the expression:

   \[
   \frac{\sqrt{3 \cdot 4 + 6}}{-3 \cdot 4} = \frac{\sqrt{12 + 6}}{-12} = \frac{\sqrt{18}}{-12}
   \]

2. **Simplify the Square Root:**
   \[
   \frac{\sqrt{18}}{-12} = \frac{3\sqrt{2}}{-12} = -\frac{\sqrt{2}}{4}
   \]

So, the limit is:

\[
\lim_{{x \to 4}} \frac{\sqrt{3x + 6}}{-3x} = -\frac{\sqrt{2}}{4}
\]

Would you like more details or have any questions?

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Here are some related questions to deepen understanding:

1. How would the solution change if the limit approached \( x = -4 \) instead?
2. What if the denominator was \( 3x \) instead of \( -3x \)?
3. How can we apply L'Hôpital's Rule to a similar problem?
4. How does the result differ if the expression inside the square root changes?
5. What happens if the limit approaches infinity?

**Tip:** Always try direct substitution first for limits. If it leads to an indeterminate form, then consider alternative methods.

Evaluate the limit: lim(x → 4) (sqrt(3x + 6) / -3x)

Explore how to evaluate the limit of sqrt(3x + 6) divided by -3x as x approaches 4. This problem demonstrates the use of direct substitution and square root simplification techniques, suitable for high school students studying calculus.

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