Let's solve the given limit problem step by step. The problem is:

$\lim_{{x \to 1}} \frac{\frac{1}{{8 - 3x}} - \frac{1}{{11}}}{{\frac{1}{{6 - 3x}} - \frac{1}{{9}}}}$

Step 1: Simplify the numerator and denominator

First, let's work on the numerator: $\text{Numerator} = \frac{1}{{8 - 3x}} - \frac{1}{{11}}$ This can be rewritten using a common denominator: $\text{Numerator} = \frac{11 - (8 - 3x)}{{11(8 - 3x)}} = \frac{3x - 8 + 11}{{11(8 - 3x)}} = \frac{3x + 3}{{11(8 - 3x)}}$ $\text{Numerator} = \frac{3(x + 1)}{{11(8 - 3x)}}$

Now, let's simplify the denominator: $\text{Denominator} = \frac{1}{{6 - 3x}} - \frac{1}{{9}}$ This can also be rewritten using a common denominator: $\text{Denominator} = \frac{9 - (6 - 3x)}{{9(6 - 3x)}} = \frac{3x - 6 + 9}{{9(6 - 3x)}} = \frac{3x + 3}{{9(6 - 3x)}}$ $\text{Denominator} = \frac{3(x + 1)}{{9(6 - 3x)}}$

Step 2: Combine the results

Now, the entire expression becomes: $\lim_{{x \to 1}} \frac{\frac{3(x + 1)}{{11(8 - 3x)}}}{{\frac{3(x + 1)}{{9(6 - 3x)}}}}$

We can simplify by canceling out the common factor $3(x + 1)$ in both the numerator and denominator: $\lim_{{x \to 1}} \frac{9(6 - 3x)}{11(8 - 3x)}$

Step 3: Substitute $x = 1$

Substitute $x = 1$ into the simplified expression: $\frac{9(6 - 3(1))}{11(8 - 3(1))} = \frac{9(6 - 3)}{11(8 - 3)} = \frac{9 \times 3}{11 \times 5} = \frac{27}{55}$

Thus, the exact answer is: $\boxed{\frac{27}{55}}$

Would you like further details on any of these steps, or do you have any questions?

Here are 5 related questions for further practice:

1. How would you solve $\lim_{{x \to 2}} \frac{\frac{1}{{5 - 2x}} - \frac{1}{{7}}}{{\frac{1}{{3 - x}} - \frac{1}{{4}}}}$?
2. What is the significance of common denominators in simplifying limit expressions?
3. How do you apply L'Hôpital's Rule to solve indeterminate forms in limits?
4. Can you find the limit $\lim_{{x \to 0}} \frac{\sin(2x)}{x}$?
5. What techniques can be used to evaluate limits that result in $\frac{0}{0}$ indeterminate forms?

Tip: Always try to simplify complex fractions in limits by finding a common denominator or factoring common terms before substituting the limit value. This often avoids indeterminate forms and simplifies the evaluation process.