The problem asks to evaluate the limit:

$\lim_{{x \to 2}} \frac{x^2 - x - 6}{x + 2}$

Step-by-Step Solution:

1. Factor the numerator:
   The expression in the numerator $x^2 - x - 6$ can be factored. We look for two numbers that multiply to $-6$ and add to $-1$. These numbers are $-3$ and $2$. Hence, we can factor the numerator as:

   $x^2 - x - 6 = (x - 3)(x + 2)$

2. Simplify the expression:
   Now the limit expression becomes:

   $\lim_{{x \to 2}} \frac{(x - 3)(x + 2)}{x + 2}$

   Since $x + 2$ appears in both the numerator and the denominator, they cancel out (as long as $x \neq -2$):

   $\lim_{{x \to 2}} (x - 3)$

3. Evaluate the limit:
   Now we can directly substitute $x = 2$ into the simplified expression:

   $2 - 3 = -1$

Final Answer:

$\lim_{{x \to 2}} \frac{x^2 - x - 6}{x + 2} = -1$

Let me know if you need further clarifications or additional details!

Here are five related questions to enhance understanding:

1. What would be the limit if $x$ approaches $-2$ in the same expression?
2. How do you handle limits that result in indeterminate forms like $\frac{0}{0}$?
3. What techniques can be used if factoring does not work for simplifying limits?
4. How does L'Hôpital's Rule apply to this type of limit problem?
5. What is the significance of removable discontinuities in the context of limits?

Tip: Always try factoring or simplifying expressions before using advanced limit techniques like L'Hôpital’s Rule!