1. **State the problem:** Find the limit as $t$ approaches 2 of the expression $$\frac{t^2 - 4}{2t^2 - 13t + 18}$$. 2. **Recall the limit rule:** If direct substitution results in an indeterminate form like $\frac{0}{0}$, try to simplify the expression by factoring and canceling common factors. 3. **Check direct substitution:** Substitute $t=2$: $$\frac{2^2 - 4}{2(2)^2 - 13(2) + 18} = \frac{4 - 4}{8 - 26 + 18} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator and denominator:** - Numerator: $t^2 - 4 = (t - 2)(t + 2)$ - Denominator: $2t^2 - 13t + 18$ To factor the denominator, find two numbers that multiply to $2 \times 18 = 36$ and add to $-13$. These are $-9$ and $-4$. Rewrite denominator: $$2t^2 - 9t - 4t + 18 = (2t^2 - 9t) - (4t - 18) = t(2t - 9) - 2(2t - 9) = (t - 2)(2t - 9)$$ 5. **Rewrite the expression:** $$\frac{(t - 2)(t + 2)}{(t - 2)(2t - 9)}$$ 6. **Cancel common factor $(t - 2)$:** $$\frac{\cancel{(t - 2)}(t + 2)}{\cancel{(t - 2)}(2t - 9)} = \frac{t + 2}{2t - 9}$$ 7. **Evaluate the limit by direct substitution:** $$\frac{2 + 2}{2(2) - 9} = \frac{4}{4 - 9} = \frac{4}{-5} = -\frac{4}{5}$$ **Final answer:** $$\lim_{t \to 2} \frac{t^2 - 4}{2t^2 - 13t + 18} = -\frac{4}{5}$$