To solve the limit

$\lim_{{x \to 4}} \frac{x^4 - 256}{x - 4},$

we first need to recognize that directly substituting $x = 4$ into the expression leads to an indeterminate form $\frac{0}{0}$. To resolve this, we can factor the numerator.

Step-by-Step Solution

1. Factor the numerator: Notice that $x^4 - 256$ is a difference of squares:

   $x^4 - 256 = (x^2 - 16)(x^2 + 16).$

   Then, we can factor $x^2 - 16$ further as it is also a difference of squares:

   $x^2 - 16 = (x - 4)(x + 4).$

   So, the full factorization of the numerator $x^4 - 256$ is:

   $x^4 - 256 = (x - 4)(x + 4)(x^2 + 16).$

2. Rewrite the limit: Substitute the factored form of the numerator into the limit expression:

   $\lim_{{x \to 4}} \frac{(x - 4)(x + 4)(x^2 + 16)}{x - 4}.$

3. Simplify the expression: Cancel the common factor $(x - 4)$ in the numerator and denominator:

   $\lim_{{x \to 4}} (x + 4)(x^2 + 16).$

4. Evaluate the limit: Substitute $x = 4$ into the simplified expression:

   $(4 + 4)(4^2 + 16) = 8 \cdot (16 + 16) = 8 \cdot 32 = 256.$

Thus, the limit is

$\lim_{{x \to 4}} \frac{x^4 - 256}{x - 4} = 256.$

Would you like more details on any part of this solution or have any questions?

Related Questions

1. What is the limit $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$?
2. How do you find the limit of a rational function where direct substitution results in an indeterminate form?
3. How can the difference of squares formula be used to simplify limits?
4. What is L'Hopital's Rule and when is it applicable for finding limits?
5. How can polynomial division be used to find limits involving polynomials?

Tip: When faced with indeterminate forms like $\frac{0}{0}$, consider factoring, using L'Hopital's rule, or polynomial division to simplify the expression before evaluating the limit.