1. **Problem:** Find the limit $\lim_{x\to 3}\frac{x^2-9}{x-3}$.\n2. **Formula and rule:** When a direct substitution gives $\frac{0}{0}$, factor the numerator and cancel the common factor. This is allowed only for values of $x$ near the limit point, not at the point itself.\n3. Factor the numerator: $$x^2-9=(x-3)(x+3).$$\n4. Substitute that into the expression: $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}.$$\n5. Cancel the common factor carefully: $$\frac{\cancel{x-3}(x+3)}{\cancel{x-3}}=x+3.$$\n6. Now evaluate the simpler expression at $x=3$: $$\lim_{x\to 3}(x+3)=3+3=6.$$\n7. **Final answer:** $6$.