1. State the problem: Find $\lim_{x\to 3}\dfrac{x^2-9}{x-3}$.\n\n2. Identify a removable discontinuity: Since $x^2-9=(x-3)(x+3)$, the fraction becomes simpler and the limit can be found by canceling $(x-3)$.\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): $$\lim_{x\to 3}\frac{\cancel{(x-3)}(x+3)}{\cancel{(x-3)}}=\lim_{x\to 3}(x+3).$$\n\n6. Evaluate by direct substitution: $$\lim_{x\to 3}(x+3)=3+3=6.$$\n\n7. Final answer: The limit is $6$.\n