Limit Evaluation 0C1Dd5

1. **State the problem:** We want to evaluate the limit $$\lim_{h \to 0} \frac{\frac{1}{(h + 3)^2} - \frac{1}{9}}{h}$$ 2. **Recall the formula:** This is a difference quotient form, often used to find derivatives. 3. **Rewrite the expression:** $$\frac{\frac{1}{(h + 3)^2} - \frac{1}{9}}{h} = \frac{\frac{9 - (h + 3)^2}{9 (h + 3)^2}}{h}$$ 4. **Simplify the numerator inside the fraction:** $$(h + 3)^2 = h^2 + 6h + 9$$ So, $$9 - (h + 3)^2 = 9 - (h^2 + 6h + 9) = -h^2 - 6h$$ 5. **Substitute back:** $$\frac{\frac{-h^2 - 6h}{9 (h + 3)^2}}{h} = \frac{-h^2 - 6h}{9 (h + 3)^2 \cdot h}$$ 6. **Factor numerator:** $$-h^2 - 6h = -h(h + 6)$$ 7. **Cancel common factor $h$ in numerator and denominator:** $$\frac{-\cancel{h}(h + 6)}{9 (h + 3)^2 \cdot \cancel{h}} = \frac{-(h + 6)}{9 (h + 3)^2}$$ 8. **Evaluate the limit as $h \to 0$:** $$\lim_{h \to 0} \frac{-(h + 6)}{9 (h + 3)^2} = \frac{-(0 + 6)}{9 (0 + 3)^2} = \frac{-6}{9 \cdot 9} = \frac{-6}{81} = -\frac{2}{27}$$ **Final answer:** $$\boxed{-\frac{2}{27}}$$