The limit expression provided is:

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

Step 1: Simplify the Expression

First, we can factor the denominator. The quadratic $x^2 - x - 2$ can be factored as:

$x^2 - x - 2 = (x - 2)(x + 1)$

Thus, the limit becomes:

$\lim_{{x \to 2}} \frac{{x - 2}}{{(x - 2)(x + 1)}}$

We can now cancel the $x - 2$ terms from the numerator and denominator, which gives:

$\lim_{{x \to 2}} \frac{1}{{x + 1}}$

Step 2: Substitute $x = 2$

Now that we have simplified the expression, we can directly substitute $x = 2$ into the remaining expression:

$\frac{1}{2 + 1} = \frac{1}{3}$

Thus, the limit is:

$\boxed{\frac{1}{3}}$

Table of Values to Support the Limit

We can construct a table of values around $x = 2$ to verify the behavior of the function:

As $x$ approaches 2 from both sides, the function value approaches $\frac{1}{3}$, which confirms our earlier result.

Would you like further clarification or details? Here are some related questions to explore further:

1. How do you factor quadratics like $x^2 - x - 2$?
2. What happens when the $x - 2$ terms don't cancel in a limit problem?
3. Why is direct substitution not possible before canceling terms?
4. How do you apply L'Hôpital's Rule for indeterminate forms?
5. Can you explain what an indeterminate form is in more detail?

Tip: Always try factoring first when you encounter quadratic expressions in limits—it can often simplify the process!