The problem is asking you to evaluate the following limit:

$\lim_{{x \to 1}} \frac{{x - 6}}{{x^2 - 3x - 18}}$

Additionally, you're asked to create a table of values for the function $f(x) = \frac{{x - 6}}{{x^2 - 3x - 18}}$ for values of $x$ close to 1. The function is discontinuous at $x = 1$, so estimating the limit numerically is a good approach.

Let's break this down:

1. Factor the denominator: The denominator is $x^2 - 3x - 18$, which we can factor: $x^2 - 3x - 18 = (x - 6)(x + 3)$

   Now, the function becomes: $f(x) = \frac{{x - 6}}{{(x - 6)(x + 3)}}$ For $x \neq 6$, this simplifies to: $f(x) = \frac{1}{{x + 3}}$ But note that this simplification holds only when $x \neq 6$, so you can use this function to fill in the table for $x$ values near 1.

2. Create the table: I'll now calculate the function $f(x) = \frac{1}{x+3}$ for the given $x$ values (0.9, 0.99, 0.999, 1.001, 1.01, and 1.1).

Let me calculate these values and fill in the table for you.Here are the values for the function $f(x) = \frac{1}{x + 3}$, rounded to four decimal places:

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