Lim_(x->3) (x2-9)/(x2+2x-3)

If we substitute $x=-3$ into the expression, we get:
Numerator: $(-3)^2 - 9 = 9 - 9 = 0$
Denominator: $(-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0$
Since we have the indeterminate form $\frac{0}{0}$, we can either factor the expression or use L'Hôpital's Rule. Let's use factoring.

The numerator is a difference of squares:

The denominator is a quadratic trinomial:

We need two numbers that multiply to $-3$ and add to 2. These numbers are 3 and $-1$.

Since $x \to -3$, $x$ is approaching $-3$ but is not equal to $-3$. Therefore, $x+3 \neq 0$, and we can cancel the $(x+3)$ term from the numerator and denominator.

Now, substitute $x=-3$ into the simplified expression:

The limit exists and is: