To solve this problem, we need to determine whether the given function $f(x)$ is continuous at $a = 8$.

Step-by-Step Process:

1. Check if $f(8)$ is defined:

   • From the piecewise function, we see that: \frac{x^2 - 64}{x - 8}, & \text{if } x \neq 8 \\ 6, & \text{if } x = 8 \end{cases}$$
   • For $x = 8$, $f(8) = 6$. So, $f(8)$ is defined.

2. Check if $\lim_{x \to 8} f(x)$ exists:

   • For $x \neq 8$, we need to simplify the expression $\frac{x^2 - 64}{x - 8}$: $\frac{x^2 - 64}{x - 8} = \frac{(x - 8)(x + 8)}{x - 8} = x + 8 \quad \text{for } x \neq 8.$
   • So, $\lim_{x \to 8} f(x) = \lim_{x \to 8} (x + 8) = 8 + 8 = 16$.

3. Check if $\lim_{x \to 8} f(x) = f(8)$:

   • From the previous step, $\lim_{x \to 8} f(x) = 16$, but $f(8) = 6$. These values are not equal.

Conclusion:

Since $\lim_{x \to 8} f(x) \neq f(8)$, the function is not continuous at $x = 8$. This matches option B: "The function is not continuous at $a = 8$ because although $f(8) = 6$ is defined and $\lim_{x \to 8} f(x)$ exists, $f(8) \neq \lim_{x \to 8} f(x)$."

Would you like further details or have any questions?

Here are five related questions to expand your understanding:

1. What are the conditions for a function to be continuous at a point?
2. How would the function change if $f(8)$ were defined as 16 instead of 6?
3. Can a function be continuous at a point if the limit does not exist?
4. How do we handle discontinuities in real-world applications of functions?
5. How does the concept of one-sided limits play into continuity at a point?

Tip: Simplifying piecewise functions to identify limits is key when dealing with continuity problems.