1. **State the problem.** We need to evaluate $\lim_{x\to 3}\frac{x^2-9}{x-3}$. 2. **Use the factoring formula.** The numerator is a difference of squares: $$x^2-9=x^2-3^2=(x-3)(x+3).$$ So the limit becomes $$\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}.$$ 3. **Cancel the common factor.** For $x\ne 3$, we can simplify the fraction: $$\frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}}=x+3.$$ This is allowed because we are finding the limit near $x=3$, not plugging in $x=3$ inside the canceled expression. 4. **Substitute the value into the simplified expression.** Now evaluate the simpler function $x+3$ at $x=3$: $$3+3=6.$$ 5. **Final answer.** $$\lim_{x\to 3}\frac{x^2-9}{x-3}=6.$$