The expression you've provided is:

$\lim_{{x \to -8^-}} (x + 12) \frac{|x + 8|}{x + 8}$

Let's break it down step-by-step:

Step 1: Behavior of $(x + 12)$

As $x$ approaches $-8$ from the left ($x \to -8^{-}$): $x + 12 \to -8 + 12 = 4$

Step 2: Behavior of $\frac{|x + 8|}{x + 8}$

• The term $|x + 8|$ is the absolute value of $x + 8$. When $x$ approaches $-8$ from the left ($x \to -8^{-}$), $x + 8$ is negative because $x$ is slightly less than $-8$.
• Therefore, $|x + 8| = -(x + 8)$, making the fraction: $\frac{|x + 8|}{x + 8} = \frac{-(x + 8)}{x + 8} = -1$

Step 3: Combine the results

Now that we know: $(x + 12) \to 4 \quad \text{and} \quad \frac{|x + 8|}{x + 8} \to -1$ The product becomes: $4 \times (-1) = -4$

Final Answer:

$\lim_{{x \to -8^{-}}} (x + 12) \frac{|x + 8|}{x + 8} = -4$

Would you like further explanation or details?

Here are 5 related questions to expand the topic:

1. What happens if the limit approaches $-8^+$ (from the right)?
2. How do absolute value functions affect limits in general?
3. Can we use L'Hopital's rule for this kind of limit?
4. How do discontinuities like absolute value impact the behavior of limits?
5. What is the geometric interpretation of such limits?

Tip: When dealing with limits involving absolute values, always consider the behavior of the function as you approach from both sides to handle discontinuities correctly.