The given problem requires finding the limit:

$\lim_{h \to 0} \frac{\sqrt[3]{27 + h} - 3}{h}.$

Solution:

Step 1: Recognize the indeterminate form

As $h \to 0$, the numerator becomes $\sqrt[3]{27} - 3 = 3 - 3 = 0$ and the denominator is also $h = 0$. Therefore, this is an indeterminate form of the type $\frac{0}{0}$, and we need to resolve it.

Step 2: Use the expansion of a cube root

The cube root function $\sqrt[3]{x}$ can be expanded as: [ \sqrt[3]{27 + h} = \sqrt[3]{27} + \frac{1}{3 \cdot \sqrt[3]{27^2}} h + \mathcal{O}(h^2), ] where $\sqrt[3]{27} = 3$.

Step 3: Substitute the expansion into the limit

Substitute $\sqrt[3]{27 + h} = 3 + \frac{h}{27} + \mathcal{O}(h^2)$ into the numerator: $\frac{\sqrt[3]{27 + h} - 3}{h} = \frac{\left(3 + \frac{h}{27} + \mathcal{O}(h^2)\right) - 3}{h}.$

Simplify: $\frac{\sqrt[3]{27 + h} - 3}{h} = \frac{\frac{h}{27} + \mathcal{O}(h^2)}{h}.$

Step 4: Simplify further

Divide through by $h$: $\frac{\sqrt[3]{27 + h} - 3}{h} = \frac{1}{27} + \mathcal{O}(h).$

Step 5: Take the limit as $h \to 0$

As $h \to 0$, the higher-order term $\mathcal{O}(h)$ vanishes. Thus: $\lim_{h \to 0} \frac{\sqrt[3]{27 + h} - 3}{h} = \frac{1}{27}.$

Final Answer:

$\boxed{\frac{1}{27}}$

Let me know if you want further clarification or details about this problem! Here are five questions to deepen understanding:

1. What other methods could you use to evaluate this limit (e.g., L'Hopital's Rule)?
2. How does the expansion for $\sqrt[3]{x}$ help in simplifying the limit?
3. Can you generalize this technique for other roots, such as square roots?
4. What would happen if $h$ were not approaching $0$?
5. Why is the higher-order term $\mathcal{O}(h^2)$ negligible in this limit?

Tip: For indeterminate forms like $\frac{0}{0}$, using expansions or derivatives often simplifies the evaluation.