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We are given the following limit problem:

\[
16 \cdot \lim_{x \to 4} \frac{\sqrt{x + 5} - \sqrt{x - 1}}{\sqrt{x + 12} - \sqrt{x - 2}}
\]

### Step 1: Direct Substitution
First, let's try to directly substitute \( x = 4 \) into the expression to check for any indeterminate forms:

\[
\frac{\sqrt{4 + 5} - \sqrt{4 - 1}}{\sqrt{4 + 12} - \sqrt{4 - 2}} = \frac{\sqrt{9} - \sqrt{3}}{\sqrt{16} - \sqrt{2}} = \frac{3 - \sqrt{3}}{4 - \sqrt{2}}
\]

This is not an indeterminate form like \( \frac{0}{0} \), so we can compute the limit directly.

### Step 2: Simplifying the expression
We already know from substitution that:
- \( \sqrt{4 + 5} = \sqrt{9} = 3 \)
- \( \sqrt{4 - 1} = \sqrt{3} \)
- \( \sqrt{4 + 12} = \sqrt{16} = 4 \)
- \( \sqrt{4 - 2} = \sqrt{2} \)

So the expression becomes:

\[
16 \cdot \frac{3 - \sqrt{3}}{4 - \sqrt{2}}
\]

This is already simplified, so we can compute the numerical value next.

### Step 3: Approximate the expression
Let me calculate the numerical value of this expression.The value of the given limit expression is approximately:

\[
16 \cdot \frac{3 - \sqrt{3}}{4 - \sqrt{2}} \approx 7.846
\]

Would you like further details on the process, or do you have any additional questions?

Here are 5 related questions to further explore the topic:

1. How would the limit change if the expression involved higher powers of \(x\)?
2. What happens to the expression as \( x \to 0 \)?
3. Can L'Hôpital's Rule be applied to this problem, and if so, how?
4. How can we rationalize the denominator to simplify similar problems?
5. What role do indeterminate forms play in determining the behavior of limits?

**Tip:** Rationalizing expressions with square roots is a common technique for simplifying complex limits. It helps avoid indeterminate forms like \( \frac{0}{0} \).

16 * lim (x -> 4) (sqrt(x + 5) - sqrt(x - 1)) / (sqrt(x + 12) - sqrt(x - 2))

Explore how to solve the limit problem involving square roots and direct substitution as x approaches 4. This problem demonstrates key techniques for evaluating limits with roots. Suitable for high school students in Grades 11-12.

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