The problem provided is:

$\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$

To solve this limit, we notice that directly substituting $x = 4$ leads to an indeterminate form of $\frac{0}{0}$. This suggests that the expression can be factored or simplified.

Step 1: Factor the numerator

The numerator $x^2 - 16$ is a difference of squares:

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

Step 2: Simplify the expression

Substitute the factored form into the original limit expression:

$\lim_{x \to 4} \frac{(x - 4)(x + 4)}{x - 4}$

Now, cancel out the common factor $x - 4$ from the numerator and denominator:

$\lim_{x \to 4} (x + 4)$

Step 3: Evaluate the limit

Now that the expression is simplified, substitute $x = 4$:

$4 + 4 = 8$

Thus, the value of the limit is:

$\boxed{8}$

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

Related Questions:

1. What would the limit be if the expression were $\lim_{x \to -4} \frac{x^2 - 16}{x + 4}$?
2. How do you apply L'Hopital's Rule to solve indeterminate forms like $\frac{0}{0}$?
3. What is the significance of factoring in solving limits algebraically?
4. Can this type of limit be solved using a graphical approach? If so, how?
5. What are the conditions under which we cannot cancel terms directly in limit problems?

Tip:

Always check for common factors in the numerator and denominator when encountering indeterminate forms such as $\frac{0}{0}$. This often simplifies the limit problem.