1. State the problem: Find the limit $$\lim_{x\to 3}\frac{x^2-9}{x-3}.$$ 2. Factor the numerator using difference of squares: $$x^2-9=x^2-3^2=(x-3)(x+3).$$ 3. Substitute the factorization into the fraction: $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}.$$ 4. Cancel the common factor (showing the canceled terms): $$\frac{(x-3)(x+3)}{\cancel{x-3}}=\cancel{\frac{x-3}{x-3}}(x+3).$$ 5. After cancellation, simplify the expression: $$=x+3.$$ 6. Evaluate the limit by plugging in $x=3$: $$\lim_{x\to 3}(x+3)=3+3=6.$$ 7. Final answer: $$\lim_{x\to 3}\frac{x^2-9}{x-3}=6.$$