We are given the following limit:

$\lim_{x \to 3} \frac{x^2 - 9x + 18}{x^2 - 10x + 21}$

Step 1: Factor both the numerator and denominator

We can try factoring the quadratic expressions in both the numerator and the denominator.

Numerator:

$x^2 - 9x + 18 = (x - 3)(x - 6)$

Denominator:

$x^2 - 10x + 21 = (x - 3)(x - 7)$

Step 2: Simplify the expression

Now that we have factored both the numerator and denominator, we can rewrite the limit as:

$\lim_{x \to 3} \frac{(x - 3)(x - 6)}{(x - 3)(x - 7)}$

We can cancel the common factor $(x - 3)$ from both the numerator and denominator, assuming $x \neq 3$. This simplifies the expression to:

$\lim_{x \to 3} \frac{x - 6}{x - 7}$

Step 3: Substitute $x = 3$

Now, substitute $x = 3$ into the simplified expression:

$\frac{3 - 6}{3 - 7} = \frac{-3}{-4} = \frac{3}{4}$

Conclusion:

The limit exists, and its value is $\frac{3}{4}$.

Would you like more details or further explanation on this? Here are some related questions:

1. How do you factor quadratic equations?
2. What is L'Hôpital's rule and when should it be applied in limits?
3. What does it mean when a limit does not exist?
4. What are common techniques for simplifying limits with indeterminate forms?
5. How does canceling terms work in simplifying limits?

Tip: Always attempt factoring first when evaluating limits of rational functions, as it often simplifies the problem.