1. State the problem: Evaluate the limit $\n\lim_{x\to 3} \frac{x^2-9}{x-3}$.\n\n2. Identify a common factor to cancel.\n\n3. Factor the numerator: $$x^2-9=(x-3)(x+3).$$\n\n4. Substitute the factorization into the limit: $$\lim_{x\to 3} \frac{(x-3)(x+3)}{x-3}.$$\n\n5. Cancel the common factor (show the canceled step clearly): $$\lim_{x\to 3} \frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}}=\lim_{x\to 3} (x+3).$$\n\n6. Evaluate by direct substitution: $$\lim_{x\to 3} (x+3)=3+3=6.$$\n\n7. Final answer: $6$.\n