To compute the limit $\lim_{x \to 4} \frac{x^2 - 16}{x^2 - 2x - 8}$, let's follow these steps:

1. Substitute $x = 4$ directly into the expression: $\frac{4^2 - 16}{4^2 - 2(4) - 8} = \frac{16 - 16}{16 - 8 - 8} = \frac{0}{0}$ This results in an indeterminate form $\frac{0}{0}$.

2. Factorize the numerator and the denominator: $\text{Numerator: } x^2 - 16 = (x - 4)(x + 4)$ $\text{Denominator: } x^2 - 2x - 8 = (x - 4)(x + 2)$

3. Simplify the expression by canceling out the common factor $(x - 4)$: $\frac{(x - 4)(x + 4)}{(x - 4)(x + 2)} = \frac{x + 4}{x + 2}, \text{ for } x \neq 4$

4. Evaluate the limit by substituting $x = 4$ into the simplified expression: $\lim_{x \to 4} \frac{x + 4}{x + 2} = \frac{4 + 4}{4 + 2} = \frac{8}{6} = \frac{4}{3}$

Thus, the limit is: $\lim_{x \to 4} \frac{x^2 - 16}{x^2 - 2x - 8} = \frac{4}{3}$

Would you like more details or have any other questions?

Here are some questions you might want to ask:

1. How do you find limits involving indeterminate forms?
2. What are some common techniques for simplifying limits?
3. How do you find the derivatives of functions involving roots?
4. How do you solve limits involving exponential functions?
5. How do you identify critical points of a function?

Tip: When evaluating limits that result in an indeterminate form, consider factoring, rationalizing, or using L'Hôpital's rule to simplify the expression.