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To solve the limit using continuity, we can substitute the value of \( x \) directly into the function, provided that the function is continuous at that point. The given limit is:

\[
\lim_{x \to 16} \frac{20 + \sqrt{x}}{\sqrt{20 + x}}
\]

Since both the numerator and the denominator involve square root functions, which are continuous everywhere they are defined (i.e., for non-negative values under the square root), and since the point \( x = 16 \) is within the domain, we can substitute \( x = 16 \) directly into the expression.

Substituting \( x = 16 \) gives:

\[
\frac{20 + \sqrt{16}}{\sqrt{20 + 16}} = \frac{20 + 4}{\sqrt{36}} = \frac{24}{6} = 4
\]

Thus, the limit is:

\[
\lim_{x \to 16} \frac{20 + \sqrt{x}}{\sqrt{20 + x}} = 4
\]

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

Here are 5 related questions that could help expand your understanding:

1. What is the definition of continuity, and how does it apply to evaluating limits?
2. How would the limit change if the square root function had a different argument, such as \( \sqrt{x + 1} \)?
3. What happens to the limit if the value of \( x \) approaches a point where the function is not continuous?
4. How would you solve this limit if the expression inside the square root was negative?
5. Can you think of another method to evaluate this limit if substitution were not possible?

**Tip:** When dealing with limits involving square roots, always check the domain of the function to ensure that the point of interest is within the domain before substituting directly.

Use continuity to evaluate the limit \( \lim_{x \to 16} \frac{20 + \sqrt{x}}{\sqrt{20 + x}} \).

Learn how to evaluate the limit of the function \( \frac{20 + \sqrt{x}}{\sqrt{20 + x}} \) as \( x \) approaches 16 using the concept of continuity. This problem involves understanding limits, continuity, and square roots, making it suitable for students in Grades 10-12.

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