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The given problem asks to evaluate the limit:

\[
\lim_{x \to 16} \frac{\sqrt{x} - 4}{x - 16}
\]

To solve this limit, we can start by rationalizing the numerator, since directly substituting \( x = 16 \) results in an indeterminate form \( \frac{0}{0} \).

### Steps:

1. **Multiply by the conjugate**: To rationalize the numerator, multiply both the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x} + 4 \).

\[
\frac{\sqrt{x} - 4}{x - 16} \times \frac{\sqrt{x} + 4}{\sqrt{x} + 4} = \frac{(\sqrt{x} - 4)(\sqrt{x} + 4)}{(x - 16)(\sqrt{x} + 4)}
\]

2. **Simplify the expression**: The numerator now becomes a difference of squares:

\[
(\sqrt{x} - 4)(\sqrt{x} + 4) = x - 16
\]

So the expression simplifies to:

\[
\frac{x - 16}{(x - 16)(\sqrt{x} + 4)} = \frac{1}{\sqrt{x} + 4}
\]

3. **Evaluate the limit**: Now, substitute \( x = 16 \) into the simplified expression:

\[
\lim_{x \to 16} \frac{1}{\sqrt{16} + 4} = \frac{1}{4 + 4} = \frac{1}{8}
\]

### Final Answer:
\[
\boxed{\frac{1}{8}}
\]

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

### Related Questions:
1. What is the conjugate and how is it used to simplify expressions?
2. How does rationalizing the numerator help resolve limits involving radicals?
3. What are the conditions under which a limit results in the indeterminate form \( \frac{0}{0} \)?
4. Can limits be solved using methods other than rationalizing, such as L'Hôpital's Rule?
5. What are other types of indeterminate forms and how can they be resolved?

### Tip:
When dealing with square roots in limits, rationalizing the numerator or denominator is a powerful technique to eliminate radicals and simplify the expression for evaluation.

Evaluate the limit by rationalizing the numerator: lim(x → 16) (√x - 4) / (x - 16). Simplify any fractions and radicals in your answer.

Learn how to evaluate the limit lim(x → 16) (√x - 4) / (x - 16) by rationalizing the numerator. This detailed solution covers rationalization techniques and the difference of squares formula, ideal for students in Grades 11-12.

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