The given expression involves a limit:

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

Step 1: Factor the numerator

We notice that the numerator $x^2 - 16$ is a difference of squares, which can be factored as:

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

So the expression becomes:

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

Step 2: Simplify the expression

Notice that $4 - x$ can be rewritten as $-(x - 4)$. This simplifies the fraction to:

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

Step 3: Cancel common terms

The terms $(x - 4)$ in the numerator and denominator cancel out, so the expression reduces to:

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

Step 4: Evaluate the limit

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

$-(4 + 4) = -8$

Conclusion:

The value of the limit is $-8$.

Do you have any questions, or would you like further details?

Here are 5 related questions for further practice:

1. What happens to the limit if $x \to 4^-$ instead of $x \to 4^+$?
2. How do you handle limits when there's no clear factorization?
3. What is the limit of $\lim_{x \to 0^+} \frac{x^2 - 1}{1 - x}$?
4. How does the limit change if the denominator is $x - 4$ instead of $4 - x$?
5. What are some common techniques to simplify limits involving indeterminate forms?

Tip: Always check for possible factorizations when dealing with polynomial limits, as this often simplifies the problem significantly.