The problem asks you to evaluate the limit of the given function as $x$ approaches 3:

$\lim_{x \to 3} \frac{x^2 - 5x + 6}{x^2 - 6x + 9}$

Let's solve this step by step.

Step 1: Substitute $x = 3$ into the expression

Substituting $x = 3$ directly into the function:

$\frac{3^2 - 5(3) + 6}{3^2 - 6(3) + 9} = \frac{9 - 15 + 6}{9 - 18 + 9} = \frac{0}{0}$

This gives us an indeterminate form $\frac{0}{0}$, which means we need to simplify the expression further to evaluate the limit.

Step 2: Factor the numerator and the denominator

Let's factor both the numerator and the denominator.

1. Numerator: $x^2 - 5x + 6$ $x^2 - 5x + 6 = (x - 2)(x - 3)$

2. Denominator: $x^2 - 6x + 9$ $x^2 - 6x + 9 = (x - 3)(x - 3) = (x - 3)^2$

Now, the function becomes:

$\frac{(x - 2)(x - 3)}{(x - 3)^2}$

Step 3: Simplify the expression

We can cancel out the common factor $(x - 3)$ from the numerator and the denominator:

$\frac{(x - 2)(x - 3)}{(x - 3)^2} = \frac{x - 2}{x - 3} \text{ for } x \neq 3$

However, because we are taking the limit as $x$ approaches 3, the expression simplifies to:

$\lim_{x \to 3} \frac{x - 2}{x - 3}$

Step 4: Evaluate the limit

Substitute $x = 3$ into the simplified expression:

$\lim_{x \to 3} \frac{x - 2}{x - 3} = \frac{3 - 2}{3 - 3} = \frac{1}{0}$

Step 5: Interpret the result

The result $\frac{1}{0}$ suggests that the limit does not exist because dividing by zero leads to an undefined value. Since the numerator is positive and the denominator approaches zero, this indicates that the limit tends to infinity or negative infinity depending on the direction of approach.

Conclusion: The limit does not exist in a finite sense. It tends toward $+\infty$ or $-\infty$ depending on the direction of approach.

Would you like further details or have any questions?

Here are some related questions to consider:

1. How can you determine if a limit tends towards $+\infty$ or $-\infty$ in such cases?
2. What is the significance of the indeterminate form $\frac{0}{0}$ in calculus?
3. How can limits be evaluated using L'Hopital's Rule?
4. Can all rational functions be simplified by factoring, as in this example?
5. How does the concept of a removable discontinuity relate to this problem?

Tip: When faced with indeterminate forms like $\frac{0}{0}$, try simplifying the expression by factoring or using algebraic manipulation before considering advanced techniques like L'Hopital's Rule.