To evaluate the limit as $h$ approaches 0 for the expression $\frac{(h-1)^3 + 1}{h}$, we start by simplifying the expression inside the limit.

Given: $\lim_{h \to 0} \frac{(h-1)^3 + 1}{h}$

Step 1: Expand $(h-1)^3$

First, expand $(h-1)^3$: $(h-1)^3 = (h-1)(h-1)(h-1)$ Expanding this: $= (h^2 - 2h + 1)(h-1) = h^3 - 3h^2 + 3h - 1$

So, we have: $(h-1)^3 + 1 = h^3 - 3h^2 + 3h - 1 + 1 = h^3 - 3h^2 + 3h$

Step 2: Substitute back into the limit

Substituting this into the original limit expression: $\lim_{h \to 0} \frac{h^3 - 3h^2 + 3h}{h}$

Step 3: Simplify the expression

Now, simplify the fraction by dividing each term by $h$: $\frac{h^3 - 3h^2 + 3h}{h} = h^2 - 3h + 3$

Step 4: Take the limit as $h$ approaches 0

Now, evaluate the limit as $h$ approaches 0: $\lim_{h \to 0} (h^2 - 3h + 3) = 0^2 - 3(0) + 3 = 3$

Thus, the limit is: $\boxed{3}$

Would you like more details, or do you have any questions?

Here are 5 related questions to explore further:

1. What is the definition of a limit, and how is it applied in calculus?
2. How does the expansion of binomials work, and why is it useful in finding limits?
3. What happens if you apply L'Hôpital's Rule to similar limit problems?
4. How does this limit relate to the concept of derivatives in calculus?
5. What are common pitfalls when evaluating limits that involve polynomial expressions?

Tip: Always check if you can simplify the expression inside a limit before directly substituting the value of the variable approaching the limit.