1. State the problem: Find $\lim_{x\to 3}\dfrac{x^2-9}{x-3}.$ 2. Identify the form and the factorization rule: Since $x^2-9$ is a difference of squares, $$x^2-9=(x-3)(x+3).$$ 3. Substitute the factorization into the expression: $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}.$$ 4. Cancel the common factor (showing the cancellation explicitly): $$\frac{(x-3)(x+3)}{x-3}=\frac{\cancel{x-3}(x+3)}{\cancel{x-3}}.$$ 5. Simplify after cancellation: $$\frac{\cancel{x-3}(x+3)}{\cancel{x-3}}=x+3.$$ 6. Evaluate the limit by plugging in $x=3$: $$\lim_{x\to 3}(x+3)=3+3=6.$$ 7. Final answer: $6$.