Evaluating a Limit by Factoring

Factoring $x^2-9$ as $(x-3)*(x+3)$ reveals a common factor with the denominator.

Cancelling the factor $x-3$ in $\frac{x^2-9}{x-3}$ is mathematically invalid.

The original function is undefined at $x=3$, but the limit exists and equals 6.

After canceling the common factor, the simplified expression becomes $x+3$, so the limit as $x\to3$ is 6.