The limit problem you provided involves evaluating:

$\lim_{{x \to 2}} \frac{3f(x)}{g(x)}$

Given that:

• $\lim_{{x \to 2}} f(x) = 4$
• $\lim_{{x \to 2}} g(x) = -2$
• $\lim_{{x \to 2}} h(x) = 0$

We can proceed by applying the limit laws. Since both $f(x)$ and $g(x)$ have defined limits as $x \to 2$, the limit of the quotient can be computed as:

$\lim_{{x \to 2}} \frac{3f(x)}{g(x)} = \frac{3 \cdot \lim_{{x \to 2}} f(x)}{\lim_{{x \to 2}} g(x)}$

Substituting the given limits:

$= \frac{3 \cdot 4}{-2} = \frac{12}{-2} = -6$

So the solution is:

$\lim_{{x \to 2}} \frac{3f(x)}{g(x)} = -6$

Would you like more details or have any additional questions?

Here are some related questions for deeper exploration:

1. What are the properties of limits that allow for this simplification?
2. How would the solution change if $g(x)$ approached zero as $x \to 2$?
3. What happens if either $f(x)$ or $g(x)$ has an undefined limit at $x = 2$?
4. How would you apply L'Hopital's rule to similar limit problems?
5. Can you provide an example where limits involve piecewise functions?

Tip: Always check the denominator in a limit involving a quotient to avoid division by zero errors.