To simplify the expression $\frac{x^2 - 16}{x - 4}$ as $x$ approaches 4, we follow these steps:

1. Factor the numerator

Notice that $x^2 - 16$ is a difference of squares, which can be factored as: $x^2 - 16 = (x - 4)(x + 4)$

2. Rewrite the expression

Now, rewrite the original expression $\frac{x^2 - 16}{x - 4}$ using the factored form of the numerator: $\frac{x^2 - 16}{x - 4} = \frac{(x - 4)(x + 4)}{x - 4}$

3. Simplify by canceling common factors

Since $x - 4$ is a factor in both the numerator and the denominator (and $x \neq 4$ to avoid division by zero), you can cancel out the $x - 4$ terms: $\frac{(x - 4)(x + 4)}{x - 4} = x + 4$

4. Evaluate the limit as $x \to 4$

Now, as $x$ approaches 4, the simplified expression becomes: $x + 4$ Substitute $x = 4$: $4 + 4 = 8$

Final Answer:

$\lim_{x \to 4} \frac{x^2 - 16}{x - 4} = 8$

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What is the difference of squares and how is it applied in factoring?
2. Why does the $x - 4$ term cancel out, and what does it imply about limits?
3. How do you evaluate limits involving removable discontinuities?
4. What happens if $x$ were exactly equal to 4 in the original expression?
5. How does this technique apply to more complex rational expressions?

Tip: When simplifying rational expressions, always check for common factors that can be canceled before substituting values in limits.