The limit given is:

$\lim_{h \to 0} \frac{(x-h)^3 - x^3}{h}$

Steps to Evaluate the Limit:

1. Expand $(x - h)^3$ using the binomial theorem:

   $(x - h)^3 = x^3 - 3x^2h + 3xh^2 - h^3$

2. Substitute this expansion into the limit expression:

   $\lim_{h \to 0} \frac{x^3 - 3x^2h + 3xh^2 - h^3 - x^3}{h}$

3. Simplify the expression:

   Notice that $x^3$ and $-x^3$ cancel out:

   $\lim_{h \to 0} \frac{-3x^2h + 3xh^2 - h^3}{h}$

   Factor $h$ out of the numerator:

   $\lim_{h \to 0} \frac{h(-3x^2 + 3xh - h^2)}{h}$

   Cancel $h$ in the numerator and denominator:

   $\lim_{h \to 0} (-3x^2 + 3xh - h^2)$

4. Evaluate the limit as $h$ approaches 0:

   As $h \to 0$, the terms involving $h$ go to 0:

   $-3x^2 + 3x(0) - 0^2 = -3x^2$

So, the limit is:

$\boxed{-3x^2}$

Would you like more details on this process or have any other questions?

Related Questions:

1. How can this result be connected to the derivative of a function?
2. What would the limit be if the function was $(x-h)^2 - x^2$ instead of $(x-h)^3 - x^3$?
3. How can this limit be derived using the definition of the derivative?
4. What would happen if the expression involved higher powers, like $(x-h)^n - x^n$?
5. How does this approach differ from evaluating a limit using L'Hôpital's Rule?

Tip:

The limit you've just calculated is a common derivative form. Specifically, it represents the derivative of $x^3$ at a point $x$.