1. **State the problem.**\nFind the limit $\lim_{x\to 3}\dfrac{x^2-9}{x-3}$.\n\n2. **Use the standard limit idea (factor then cancel).**\nHere the expression is a $0/0$ form when $x=3$, because $x^2-9=9-9=0$ and $x-3=0$.\n\n3. **Factor the numerator.**\n$$x^2-9=(x-3)(x+3)$$\n\n4. **Substitute the factored form into the fraction.**\n$$\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}$$\n\n5. **Cancel the common factor (show the cancellation).**\n$$\dfrac{\cancel{(x-3)}(x+3)}{\cancel{(x-3)}}=x+3$$\n\n6. **Evaluate the limit by plugging in $x=3$.**\n$$\lim_{x\to 3}(x+3)=3+3=6$$\n\n7. **Final answer.**\n$\boxed{6}$\n