The given problem is:

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

Step-by-Step Solution:

Step 1: Substituting $x = 3$

When we substitute $x = 3$ directly into the expression: $\frac{x^2 - x - 6}{x^2 - 9} = \frac{3^2 - 3 - 6}{3^2 - 9} = \frac{9 - 3 - 6}{9 - 9} = \frac{0}{0}.$

This results in an indeterminate form ($\frac{0}{0}$), so we need to simplify the expression.

Step 2: Factoring the numerator and denominator

1. The numerator $x^2 - x - 6$ can be factored as: $x^2 - x - 6 = (x - 3)(x + 2).$

2. The denominator $x^2 - 9$ is a difference of squares: $x^2 - 9 = (x - 3)(x + 3).$

So, the expression becomes: $\frac{x^2 - x - 6}{x^2 - 9} = \frac{(x - 3)(x + 2)}{(x - 3)(x + 3)}.$

Step 3: Canceling the common factor

Since $x \neq 3$ (as $x \to 3$), we can cancel the common factor $x - 3$: $\frac{(x - 3)(x + 2)}{(x - 3)(x + 3)} = \frac{x + 2}{x + 3}.$

Step 4: Substituting $x = 3$

Now, substitute $x = 3$ into the simplified expression: $\frac{x + 2}{x + 3} = \frac{3 + 2}{3 + 3} = \frac{5}{6}.$

Final Answer:

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

Let me know if you'd like more details or have any further questions!

5 Related Questions:

1. What is the meaning of an indeterminate form in limits?
2. How do you factorize a quadratic expression like $x^2 - x - 6$?
3. What is the difference of squares formula, and when is it used?
4. How can we handle limits involving higher-order polynomials?
5. Can L'Hôpital's Rule be applied to this problem, and how?

Tip:

Always simplify the expression algebraically before substituting the value of $x$ in limit problems to avoid indeterminate forms like $\frac{0}{0}$.