1. **Problem:** Evaluate the limit $\lim_{x\to 3}\frac{x^2-9}{x-3}$. 2. **Formula / idea:** When direct substitution gives $\frac{0}{0}$, factor the numerator and cancel the common factor. 3. **Factor the numerator:** $$x^2-9=(x-3)(x+3)$$ 4. **Rewrite the expression:** $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}$$ 5. **Cancel the common factor:** $$\frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}}=x+3$$ 6. **Now evaluate the limit:** $$\lim_{x\to 3}(x+3)=3+3=6$$ 7. **Final answer:** $6$