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The problem is to find the value of:

\[
\lim_{{x \to 5}} \frac{\sqrt{2x - 1} - \sqrt{x + 4}}{\sqrt{x - 1} - \sqrt{3x - 11}}
\]

### Solution:

1. **Direct Substitution**:
   By substituting \( x = 5 \), we get:

   - Numerator: \(\sqrt{2 \cdot 5 - 1} - \sqrt{5 + 4} = \sqrt{10 - 1} - \sqrt{9} = \sqrt{9} - 3 = 3 - 3 = 0\).
   - Denominator: \(\sqrt{5 - 1} - \sqrt{3 \cdot 5 - 11} = \sqrt{4} - \sqrt{15 - 11} = 2 - 2 = 0\).

   This results in a \( \frac{0}{0} \) indeterminate form, so we need to simplify further.

2. **Rationalizing the Numerator and Denominator**:
   To simplify, we will rationalize both the numerator and the denominator.

   - **Numerator**:
     Multiply the numerator by the conjugate, \(\sqrt{2x - 1} + \sqrt{x + 4}\):
     \[
     \frac{\left(\sqrt{2x - 1} - \sqrt{x + 4}\right) \cdot \left(\sqrt{2x - 1} + \sqrt{x + 4}\right)}{\sqrt{2x - 1} + \sqrt{x + 4}} = \frac{(2x - 1) - (x + 4)}{\sqrt{2x - 1} + \sqrt{x + 4}} = \frac{x - 5}{\sqrt{2x - 1} + \sqrt{x + 4}}
     \]

   - **Denominator**:
     Similarly, multiply the denominator by the conjugate, \(\sqrt{x - 1} + \sqrt{3x - 11}\):
     \[
     \frac{\left(\sqrt{x - 1} - \sqrt{3x - 11}\right) \cdot \left(\sqrt{x - 1} + \sqrt{3x - 11}\right)}{\sqrt{x - 1} + \sqrt{3x - 11}} = \frac{(x - 1) - (3x - 11)}{\sqrt{x - 1} + \sqrt{3x - 11}} = \frac{-2x + 10}{\sqrt{x - 1} + \sqrt{3x - 11}} = \frac{-2(x - 5)}{\sqrt{x - 1} + \sqrt{3x - 11}}
     \]

3. **Simplify the Expression**:
   Substitute these into the original limit expression:
   \[
   \lim_{{x \to 5}} \frac{\frac{x - 5}{\sqrt{2x - 1} + \sqrt{x + 4}}}{\frac{-2(x - 5)}{\sqrt{x - 1} + \sqrt{3x - 11}}} = \lim_{{x \to 5}} \frac{x - 5}{\sqrt{2x - 1} + \sqrt{x + 4}} \cdot \frac{\sqrt{x - 1} + \sqrt{3x - 11}}{-2(x - 5)}
   \]

   Cancel out \((x - 5)\) in the numerator and denominator:
   \[
   = \lim_{{x \to 5}} \frac{1}{\sqrt{2x - 1} + \sqrt{x + 4}} \cdot \frac{\sqrt{x - 1} + \sqrt{3x - 11}}{-2}
   \]

4. **Substitute \( x = 5 \) Again**:
   Now, substitute \( x = 5 \):
   - The first term: \(\sqrt{2 \cdot 5 - 1} + \sqrt{5 + 4} = \sqrt{9} + \sqrt{9} = 3 + 3 = 6\).
   - The second term: \(\sqrt{5 - 1} + \sqrt{3 \cdot 5 - 11} = \sqrt{4} + \sqrt{4} = 2 + 2 = 4\).

   So, the limit becomes:
   \[
   \frac{1}{6} \cdot \frac{4}{-2} = \frac{1}{6} \cdot -2 = -\frac{1}{3}
   \]

### Final Answer:
\[
\lim_{{x \to 5}} \frac{\sqrt{2x - 1} - \sqrt{x + 4}}{\sqrt{x - 1} - \sqrt{3x - 11}} = -\frac{1}{3}
\]

Would you like further clarification or details on any step?

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Here are some related questions to deepen your understanding:

1. What would happen if the limit approaches a different point, such as \( x \to 4 \)?
2. How can we use L'Hôpital's Rule to solve similar indeterminate forms?
3. What are other methods to resolve indeterminate forms besides rationalizing?
4. How does the concept of conjugates apply to simplifying limits?
5. How would this approach differ if the functions were not square roots but cube roots?

**Tip:** In limits with radicals, rationalizing by multiplying with the conjugate often simplifies expressions, especially in cases involving indeterminate forms.

Find the value of lim (x -> 5) [(sqrt(2x - 1) - sqrt(x + 4)) / (sqrt(x - 1) - sqrt(3x - 11))].

Learn how to evaluate the limit as x approaches 5 for the expression [(sqrt(2x - 1) - sqrt(x + 4)) / (sqrt(x - 1) - sqrt(3x - 11))]. This step-by-step solution covers rationalization techniques and limit theorems for indeterminate forms, ideal for Grades 11-12.

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