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The problem involves finding the limit:

\[
\lim_{{x \to 9}} \frac{x^2 + 6x - 27}{x + 9}
\]

### Step 1: Substitute \( x = 9 \) directly
To see if direct substitution works, we plug \( x = 9 \) into the expression:

\[
\frac{9^2 + 6 \cdot 9 - 27}{9 + 9} = \frac{81 + 54 - 27}{18} = \frac{108}{18} = 6
\]

### Final Answer:
\[
\lim_{{x \to 9}} \frac{x^2 + 6x - 27}{x + 9} = 6
\]

Let me know if you have any questions, need further clarification, or want a deeper explanation of this problem. 

#### Related Questions
1. How would you solve the limit if direct substitution led to an indeterminate form?
2. Can this limit be solved by factoring? Why or why not?
3. What if the limit was \( x \to -9 \) instead?
4. How do you find the limit of a rational function with a factorable numerator?
5. What is the significance of limits in calculus?

#### Tip
Always check for direct substitution first in limit problems—it's the quickest method if it works!

Evaluate the limit: \( \lim_{{x \to 9}} \frac{x^2 + 6x - 27}{x + 9} \)

Learn how to evaluate the limit of \( \frac{x^2 + 6x - 27}{x + 9} \) as \( x \) approaches 9. This solution demonstrates the use of direct substitution for limit evaluation, focusing on key concepts in calculus.

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