Using L'Hôpital's rule (as the limit is of the form $0/0$ when $x \to -a$ since $(-a)^9 + a^9 = -a^9 + a^9 = 0$):
$\lim_{x \to -a} \frac{\frac{d}{dx}(x^9 + a^9)}{\frac{d}{dx}(x + a)} = \lim_{x \to -a} \frac{9x^8}{1} = 9(-a)^8 = 9a^8$.
We are given that this limit equals 9:
$9a^8 = 9 \implies a^8 = 1$.
Taking real roots: $a = \pm 1$.

Detailed analysis of options:

• $a = \pm 1$ is correct ($a^8 = 1 \implies a = \pm 1$).
• $a = \pm 9$ is incorrect.
• $a = \pm 3$ is incorrect.
• $a = \pm 8$ is incorrect.