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!