We are tasked with evaluating the limit:

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

This expression seems to have the indeterminate form $\frac{0}{0}$ when we substitute $x = 4$ directly. To resolve this, we need to simplify the expression.

Step 1: Simplify the numerator

The numerator is $-x^2 + 16$, which can be rewritten as:

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

Thus, the limit becomes:

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

Step 2: Cancel the common factor

The $(x - 4)$ terms cancel out (as long as $x \neq 4$), leaving us with:

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

Step 3: Evaluate the limit

Now, substitute $x = 4$ into the simplified expression:

$-(4 + 4) = -8$

Thus, the limit is:

$\boxed{-8}$

Identifying $f(x)$ and $c$

We are also asked to identify $f(x)$ and $c$:

• The function $f(x)$ is likely the original function $f(x) = -x^2 + 16$.
• The number $c$ is the value at which we are taking the limit, $c = 4$.

Would you like more details or have any questions? Here are 5 related questions:

1. How would the limit change if $f(x)$ was a different quadratic function?
2. What happens if we approach a different value of $c$ instead of 4?
3. How can we apply L'Hopital's Rule to evaluate limits like this?
4. What if $f(x)$ involved a higher degree polynomial, like $x^3$?
5. How do limits relate to the concept of a derivative in calculus?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ before simplifying limits involving polynomials.