1. State the problem: Find $$\lim_{x\to 3}\frac{x^2-9}{x-3}.$$\n\n2. Substitute directly (quick check): The form is $$\frac{0}{0}$$ because $$x^2-9=(x-3)(x+3).$$\n\n3. Factor the numerator: $$x^2-9=(x-3)(x+3).$$\n\n4. Rewrite the limit using the factorization: $$\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}.$$\n\n5. Cancel the common factor with an explicit cancellation step: $$\lim_{x\to 3}\frac{(x+3)\cancel{(x-3)}}{\cancel{(x-3)}}.$$\n\n6. Now evaluate by direct substitution: $$\lim_{x\to 3}(x+3)=3+3=6.$$\n\n7. Final answer: $$6.$$\n