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To solve this limit using continuity, let's evaluate it by direct substitution since the function is continuous at \( x = 4 \).

The limit is:
\[
\lim_{{x \to 4}} \frac{32 + \sqrt{x}}{\sqrt{32 + x}}
\]

1. **Substitute \( x = 4 \) directly:**
   \[
   \frac{32 + \sqrt{4}}{\sqrt{32 + 4}} = \frac{32 + 2}{\sqrt{36}}
   \]

2. **Simplify each part:**
   - \( 32 + 2 = 34 \)
   - \( \sqrt{36} = 6 \)

3. **Complete the division:**
   \[
   \frac{34}{6} = \frac{17}{3}
   \]

Thus, the result of the limit is:
\[
\lim_{{x \to 4}} \frac{32 + \sqrt{x}}{\sqrt{32 + x}} = \frac{17}{3}
\]

Would you like a more detailed explanation, or do you have any questions?

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Here are 5 related questions that build upon this concept:
1. What is continuity, and why does it allow for direct substitution in limits?
2. How do we determine if a function is continuous at a point?
3. Can limits be evaluated by continuity if there is an indeterminate form, such as \( \frac{0}{0} \)?
4. What are some alternative methods for evaluating limits when direct substitution is not possible?
5. How does the square root function affect the continuity and differentiability of a function?

**Tip:** When dealing with limits involving square roots, simplifying expressions or using direct substitution can often make the solution straightforward if the function is continuous at that point.

Use continuity to evaluate the limit: \( \lim_{{x \to 4}} \frac{32 + \sqrt{x}}{\sqrt{32 + x}} \).

Learn how to evaluate the limit \( \lim_{{x \to 4}} \frac{32 + \sqrt{x}}{\sqrt{32 + x}} \) using continuity. This problem covers key calculus concepts, including limits and the continuity theorem for evaluating limits. Suitable for high school students in Grades 11-12.

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