The problem involves solving the following limit:

$\lim_{x \to 0} \frac{f(x)^3 - g(x)}{\sqrt{g(x)} - 16}$

Given that:

$\lim_{x \to 0} f(x) = 2 \quad \text{and} \quad \lim_{x \to 0} g(x) = 0$

Let us break down the limit step by step.

Step 1: Substitution of limits at $x = 0$

First, substitute $x = 0$ into the expression to see what it yields.

• From the given, $\lim_{x \to 0} f(x) = 2$, so $f(0) = 2$.
• From the given, $\lim_{x \to 0} g(x) = 0$, so $g(0) = 0$.

Substitute $f(x) = 2$ and $g(x) = 0$ into the expression:

$\frac{f(x)^3 - g(x)}{\sqrt{g(x)} - 16} \Rightarrow \frac{2^3 - 0}{\sqrt{0} - 16} = \frac{8 - 0}{0 - 16} = \frac{8}{-16} = -\frac{1}{2}$

Step 2: Check for indeterminate form

The substitution leads to the following result:

$\frac{8}{-16} = -\frac{1}{2}$

So, we have a finite value, and no indeterminate form like $\frac{0}{0}$ arises. This means we do not need to apply any advanced limit laws such as L'Hôpital's Rule.

Step 3: Conclusion

The limit evaluates directly as:

$\lim_{x \to 0} \frac{f(x)^3 - g(x)}{\sqrt{g(x)} - 16} = -\frac{1}{2}$

Further Questions:

1. How does L'Hôpital's Rule apply to limits that result in indeterminate forms like $\frac{0}{0}$?
2. What happens if $f(x)$ and $g(x)$ had different limiting values at $x = 0$?
3. What are the general rules for applying limits to rational functions?
4. How can we interpret the behavior of functions at a point if both limits lead to the same value?
5. What is the significance of continuous functions in evaluating limits?

Tip:

Always check for indeterminate forms first. If an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ arises, then more advanced tools like L'Hôpital's Rule or algebraic simplification are often needed.