https://file.aiquickdraw.com/mathbot/file/17278830112529zoxlx3j.jpg

The image shows a mathematical limit expression. Here’s the transcription of the limit problem:

\[
\lim_{{x \to 3}} \frac{\sqrt{1 - x} - 2}{8 + x}
\]

### Solution:

1. **Substitute** \( x = 3 \) directly into the expression:
   \[
   \frac{\sqrt{1 - 3} - 2}{8 + 3} = \frac{\sqrt{-2} - 2}{11}
   \]
   This results in a square root of a negative number, which is undefined in the real number system.

### Conclusion:

Since \( \sqrt{-2} \) is imaginary, this limit does not exist within the real number system.

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

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Here are 5 related questions to expand on this:
1. What happens when the limit involves complex numbers?
2. How would you approach this limit if complex numbers were allowed?
3. Can you provide a limit problem where the denominator leads to an indeterminate form?
4. What is the process for simplifying limits involving square roots?
5. How can L'Hopital's Rule be applied when the limit results in an indeterminate form?

**Tip:** Always check for undefined or imaginary values when substituting into limit expressions, especially when dealing with square roots.

Learn how to solve the limit lim_{x -> 3} (sqrt(1 - x) - 2) / (8 + x). This problem explores limits, square roots, and the concept of undefined expressions when square roots of negative numbers appear. Ideal for students in Grades 11-12.

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