The problem in the image is to find the limit:

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

To solve this, let’s go through the steps:

1. Factor the expression in the numerator. Notice that $x^2 - 9$ is a difference of squares, which factors as: $x^2 - 9 = (x + 3)(x - 3)$

   So, the limit becomes: $\lim_{{x \to 3}} \frac{(x + 3)(x - 3)}{x - 3}$

2. Cancel the common factor $(x - 3)$ from the numerator and denominator, assuming $x \neq 3$ (since we're only approaching 3, not substituting directly): $= \lim_{{x \to 3}} (x + 3)$

3. Substitute $x = 3$ in the simplified expression: $= 3 + 3 = 6$

Answer:

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

Explanation:

The original function was undefined at $x = 3$ due to a zero in the denominator. By factoring and canceling $(x - 3)$, we removed this undefined behavior, allowing us to substitute $x = 3$ in the simplified form.

Would you like additional details on factoring, limits, or other related topics?

Here are 5 further questions related to limits and factoring:

1. What happens when we try to find the limit directly without factoring?
2. How do we handle cases where factoring isn’t possible in limit problems?
3. What is the difference between limits approaching a value and actual substitution?
4. Are there cases where limits approach infinity rather than a finite number?
5. How does factoring help in solving limits involving polynomials?

Tip: Always check for factoring opportunities in limits where substitution initially gives a form like $\frac{0}{0}$; it often simplifies the problem!