1. **State the problem.** Find the limit $\lim_{x\to 3}\frac{x^2-9}{x-3}$. 2. **Use the main rule for this type of limit.** When direct substitution gives $\frac{0}{0}$, we simplify the expression by factoring and canceling the common factor. 3. **Factor the numerator.** $$x^2-9=(x-3)(x+3)$$ So the limit becomes $$\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}$$ 4. **Cancel the common factor.** For $x\neq 3$, we can simplify: $$\frac{\cancel{x-3}(x+3)}{\cancel{x-3}}=x+3$$ So now the limit is $$\lim_{x\to 3}(x+3)$$ 5. **Substitute the value.** $$3+3=6$$ 6. **Final answer.** $$\lim_{x\to 3}\frac{x^2-9}{x-3}=6$$