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The given problem is a limit expression:

\[
\lim_{{x \to 4}} \frac{{x^3 - 64}}{{x - 4}}
\]

### Solution:

First, notice that the expression for the limit is indeterminate at \(x = 4\) (since both the numerator and denominator become zero). To resolve this, we need to simplify the expression by factoring the numerator.

The numerator \(x^3 - 64\) is a difference of cubes, and can be factored as:

\[
x^3 - 64 = (x - 4)(x^2 + 4x + 16)
\]

Now, substitute this factored form back into the original limit expression:

\[
\lim_{{x \to 4}} \frac{{(x - 4)(x^2 + 4x + 16)}}{{x - 4}}
\]

Cancel out the common factor \((x - 4)\):

\[
\lim_{{x \to 4}} (x^2 + 4x + 16)
\]

Now, substitute \(x = 4\) into the remaining expression:

\[
4^2 + 4(4) + 16 = 16 + 16 + 16 = 48
\]

Thus, the value of the limit is:

\[
\boxed{48}
\]

Would you like further details or have any questions?

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### Related Questions:
1. What happens if we use L'Hopital's Rule instead of factoring in this limit problem?
2. Can this technique be applied to other types of polynomial limits?
3. How does the difference of cubes formula work in other contexts?
4. What are other common indeterminate forms when calculating limits?
5. Can factoring always help solve limits that give \( \frac{0}{0} \)?

**Tip:** Always check for factorizations when dealing with limits involving polynomials that result in indeterminate forms.

Calculate the limit lim(x → 4) [(x^3 - 64) / (x - 4)]

Solve the limit lim(x → 4) [(x^3 - 64) / (x - 4)] using factoring techniques and the difference of cubes formula. Understand how to handle indeterminate forms and apply polynomial factoring to simplify limits.

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