The problem is asking to compute the limit of the expression:

$\lim_{{x \to -8}} \frac{x^2 + 4x - 32}{x + 8}$

Step-by-step solution:

1. Substitution: First, let's try direct substitution by plugging $x = -8$ into the expression:

   $\frac{(-8)^2 + 4(-8) - 32}{-8 + 8} = \frac{64 - 32 - 32}{0} = \frac{0}{0}$

   Since direct substitution results in the indeterminate form $\frac{0}{0}$, we need to simplify the expression further.

2. Factoring the numerator: The numerator is a quadratic expression: $x^2 + 4x - 32$. Let's factor it:

   • Find two numbers that multiply to $-32$ and add up to $4$.
   • These numbers are $8$ and $-4$.

   So, we can factor the numerator as:

   $(x + 8)(x - 4)$

3. Simplifying the expression: Now substitute the factored form back into the limit expression:

   $\lim_{{x \to -8}} \frac{(x + 8)(x - 4)}{x + 8}$

   We can cancel out the $(x + 8)$ terms from the numerator and denominator:

   $\lim_{{x \to -8}} (x - 4)$

4. Evaluate the limit: Now that the expression is simplified, we can substitute $x = -8$:

   $-8 - 4 = -12$

Thus, the limit is:

$\boxed{-12}$

Would you like any additional details or clarifications?

Here are 5 related questions:

1. How do we handle other types of indeterminate forms like $\frac{0}{0}$?
2. What are the common techniques to simplify limits involving factored forms?
3. Could L'Hopital's Rule be used in this problem, and when is it typically applied?
4. How do we solve limits involving infinity?
5. What are the different types of discontinuities that might occur in limits?

Tip: When computing limits, always attempt direct substitution first to check for indeterminate forms before simplifying the expression.