To evaluate the limit

$\lim_{{x \to 4}} \frac{\sqrt{x} - 2}{x - 4}$

we notice that direct substitution results in an indeterminate form $\frac{0}{0}$. To resolve this, we can rationalize the expression by multiplying the numerator and the denominator by the conjugate of the numerator:

$\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)}$

Simplifying the numerator:

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

So the expression becomes:

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

Now we can take the limit as $x \to 4$:

$\lim_{{x \to 4}} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$

Thus, the limit is:

$\boxed{\frac{1}{4}}$

Would you like more details on this solution or have any questions?

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Here are some related questions to deepen understanding:

1. What is the purpose of rationalizing the numerator when finding limits?
2. How does the limit change if the expression were $\frac{\sqrt{x} - 3}{x - 9}$?
3. What happens if we substitute a different number instead of 4? Does the method change?
4. Can you apply L'Hôpital's Rule to this limit, and if so, how does it compare?
5. How would the solution differ if the expression were $\frac{\sqrt{x + 1} - 2}{x - 3}$?

Tip: Rationalizing is useful for handling limits that result in indeterminate forms involving square roots.