1. State the problem: Find $\lim_{x\to 3} \dfrac{x^2-9}{x-3}$.\n\n2. Recall the key rule: If direct substitution gives an indeterminate form $0/0$, factor first.\n\n3. Factor the numerator: \n$$x^2-9=(x-3)(x+3).$$\n\n4. Substitute the factorization into the limit: \n$$\lim_{x\to 3} \frac{x^2-9}{x-3}=\lim_{x\to 3} \frac{(x-3)(x+3)}{x-3}. $$\n\n5. Cancel the common factor (showing the cancellation explicitly):\n$$\lim_{x\to 3} \frac{(x-3)(x+3)}{x-3}=\lim_{x\to 3} \frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}}. $$\n\n6. Simplify after cancellation: \n$$\lim_{x\to 3} \frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}}=\lim_{x\to 3} (x+3). $$\n\n7. Evaluate by direct substitution: \n$$\lim_{x\to 3} (x+3)=3+3=6. $$\n\n8. Final answer: $6$.\n