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The given expression is:

\[
\lim_{{x \to 4}} \frac{{\sqrt{x} - 2}}{{4 - x}}
\]

We can evaluate this limit using algebraic simplification. Let's proceed step by step.

### Step 1: Substitution
If we directly substitute \( x = 4 \) into the expression:

\[
\frac{{\sqrt{4} - 2}}{{4 - 4}} = \frac{{2 - 2}}{{0}} = \frac{0}{0}
\]

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

### Step 2: Multiply by the conjugate
To simplify the expression, we can multiply both the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{x} + 2 \):

\[
\frac{{\sqrt{x} - 2}}{{4 - x}} \times \frac{{\sqrt{x} + 2}}{{\sqrt{x} + 2}} = \frac{{(\sqrt{x} - 2)(\sqrt{x} + 2)}}{{(4 - x)(\sqrt{x} + 2)}}
\]

### Step 3: Simplify the expression
The numerator becomes:

\[
(\sqrt{x} - 2)(\sqrt{x} + 2) = x - 4
\]

So the expression simplifies to:

\[
\frac{{x - 4}}{{(4 - x)(\sqrt{x} + 2)}}
\]

### Step 4: Factor cancellation
We can factor \( 4 - x \) as \( -(x - 4) \), and the expression becomes:

\[
\frac{{x - 4}}{{-(x - 4)(\sqrt{x} + 2)}} = \frac{-1}{{\sqrt{x} + 2}}
\]

### Step 5: Substitution of \( x = 4 \)
Now, substitute \( x = 4 \) into the simplified expression:

\[
\frac{-1}{{\sqrt{4} + 2}} = \frac{-1}{{2 + 2}} = \frac{-1}{4}
\]

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

Let me know if you would like more details or have any questions!

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Here are 5 related questions:
1. How can limits involving square roots be simplified in general?
2. What is the importance of multiplying by the conjugate in limit problems?
3. Can L'Hôpital's rule be applied to this problem? If so, how?
4. What are other common indeterminate forms in calculus, besides \( \frac{0}{0} \)?
5. How does factoring help in solving limits involving rational functions?

**Tip**: Always check for indeterminate forms first before deciding which limit technique to apply!

Learn how to evaluate the limit of (sqrt(x) - 2) / (4 - x) as x approaches 4 using conjugates and simplification. This calculus problem involves dealing with an indeterminate form and is suitable for high school students in Grades 10-12.

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