1. The problem asks to find which statements are equivalent or true for the limit $$\lim_{x \to 2} \frac{x^2 + 2x - 8}{x^4 - 16}$$. 2. Start by factoring numerator and denominator: - Numerator: $$x^2 + 2x - 8 = (x + 4)(x - 2)$$ - Denominator: $$x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$$ 3. Substitute factored forms into the original limit: $$\lim_{x \to 2} \frac{(x+4)(x-2)}{(x-2)(x+2)(x^2 + 4)}$$ 4. Cancel the common factor $(x-2)$ from numerator and denominator, noting $x \to 2$ but $x \neq 2$: $$\lim_{x \to 2} \frac{x+4}{(x+2)(x^2 + 4)}$$ This matches option a. 5. Evaluate the simplified limit by substituting $x=2$: $$\frac{2+4}{(2+2)(2^2 + 4)} = \frac{6}{4 \cdot (4 + 4)} = \frac{6}{4 \cdot 8} = \frac{6}{32} = \frac{3}{16}$$ 6. Check the given statements: - a. $$\lim_{x \to 2} \frac{x+4}{(x+2)(x^2 + 4)}$$ is equivalent after simplification \(\checkmark\) - b. $$\lim_{x \to 2} \frac{(x-4)(x+2)}{(x-2)(x+2)(x^2 + 4)}$$ is not equivalent, numerator factors differ \(\times\) - c. $$\lim_{x \to 2} \frac{x^2 + 2x - 8}{x^4 - 16} = -\frac{3}{16}$$ incorrect sign \(\times\) - d. $$\lim_{x \to 2} \frac{(x+4)(x-2)}{(x^2 - 4)(x^2 + 4)}$$ denominator expands to same as original denominator as $(x^2-4)=(x-2)(x+2)$, so this is equivalent \(\checkmark\) - e. $$\lim_{x \to 2} \frac{(x+4)(x-2)}{(x-2)(x+2)(x^2 + 4)}$$ canceling $(x-2)$ yields same as a, so equivalent \(\checkmark\) - f. $$\lim_{x \to 2} \frac{x^2 + 2x - 8}{x^4 - 16} = \frac{3}{16}$$ correct value \(\checkmark\) - g. $$\lim_{x \to 2} \frac{x-4}{(x-2)(x^2 + 4)}$$ numerator differs, not equivalent \(\times\) Final answers: a, d, e, f are true/equivalent.