The limit provided in the image is:

$\lim_{t \to 3} \frac{t^3 - 27}{t^2 - 9}$

Let's simplify this step by step.

Step 1: Factor both the numerator and denominator

1. The numerator is $t^3 - 27$, which is a difference of cubes. The factoring formula for $a^3 - b^3$ is: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ Here, $t^3 - 27 = t^3 - 3^3$, so: $t^3 - 27 = (t - 3)(t^2 + 3t + 9)$

2. The denominator is $t^2 - 9$, which is a difference of squares. The factoring formula for $a^2 - b^2$ is: $a^2 - b^2 = (a - b)(a + b)$ Here, $t^2 - 9 = t^2 - 3^2$, so: $t^2 - 9 = (t - 3)(t + 3)$

Step 2: Substitute the factored forms into the limit expression

The expression now becomes: $\lim_{t \to 3} \frac{(t - 3)(t^2 + 3t + 9)}{(t - 3)(t + 3)}$

Step 3: Cancel out the common factor $(t - 3)$

As long as $t \neq 3$, we can cancel the common term $(t - 3)$ from both the numerator and the denominator: $\lim_{t \to 3} \frac{t^2 + 3t + 9}{t + 3}$

Step 4: Substitute $t = 3$ into the simplified expression

Now, substitute $t = 3$ into the remaining expression: $\frac{3^2 + 3(3) + 9}{3 + 3} = \frac{9 + 9 + 9}{6} = \frac{27}{6} = 4.5$

Final Answer:

$\lim_{t \to 3} \frac{t^3 - 27}{t^2 - 9} = 4.5$

Would you like further clarification or details? Here are some additional questions to expand on this:

1. What happens if $t = 3$ is directly substituted without factoring?
2. How can limits be used to solve real-world problems?
3. Can the difference of cubes formula be applied to other higher powers?
4. What is the importance of factoring when dealing with limits?
5. How do we approach indeterminate forms like $\frac{0}{0}$ in limits?

Tip: Always try to factor the expressions in limits before applying direct substitution to avoid indeterminate forms.