1. **Problem statement:** Find the following limits: (i) $$\lim_{x \to 3^-} \frac{x^3 - 12}{x - 3}$$ (ii) $$\lim_{x \to -2^-} \frac{x^3}{x + 2}$$ 2. **Recall the limit definition and approach:** When evaluating limits where direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, we try to simplify the expression. 3. **Evaluate (i):** Substitute $$x=3$$ directly: $$\frac{3^3 - 12}{3 - 3} = \frac{27 - 12}{0} = \frac{15}{0}$$ which is undefined but not indeterminate. Since denominator approaches 0, check the sign of numerator and denominator as $$x \to 3^-$$: - Numerator near 3: $$x^3 - 12$$ at $$x=3$$ is 15 (positive). - Denominator near 3 from left: $$x - 3$$ is slightly less than 0 (negative). So the fraction approaches $$\frac{+}{-} = -\infty$$. 4. **Evaluate (ii):** Substitute $$x = -2$$ directly: $$\frac{(-2)^3}{-2 + 2} = \frac{-8}{0}$$ undefined. Check signs as $$x \to -2^-$$: - Numerator: $$x^3$$ near $$-2$$ is slightly less than $$-8$$ (negative). - Denominator: $$x + 2$$ near $$-2$$ from left is slightly less than 0 (negative). So the fraction approaches $$\frac{-}{-} = +\infty$$. **Final answers:** (i) $$\lim_{x \to 3^-} \frac{x^3 - 12}{x - 3} = -\infty$$ (ii) $$\lim_{x \to -2^-} \frac{x^3}{x + 2} = +\infty$$