The given expression is:

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

First, note that $t^2 - 9$ can be factored as a difference of squares:

$t^2 - 9 = (t - 3)(t + 3)$

Substitute this factorization into the limit:

$\lim_{{t \to 3}} \frac{{(t - 3)(t + 3)}}{{t - 3}}$

Now, cancel the common factor of $t - 3$ from the numerator and denominator (assuming $t \neq 3$):

$\lim_{{t \to 3}} (t + 3)$

Now, substitute $t = 3$ into the remaining expression:

$t + 3 = 3 + 3 = 6$

Thus, the value of the limit is:

$\boxed{6}$

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you handle limits with indeterminate forms like $0/0$?
2. What are the steps for factoring polynomials when solving limits?
3. Can you apply L'Hopital's Rule to solve this limit? How?
4. What is the general method for solving limits involving factored expressions?
5. How does the limit change if the numerator had been $t^2 - 6$?

Tip: Always check for common factors in limits involving polynomials to simplify the expression before evaluating the limit!