The limit of the function $\frac{9}{x^3}$ as $x$ approaches negative infinity is defined as:

$\lim_{x \to -\infty} \frac{9}{x^3}$

Step-by-step process:

1. Analyze the behavior of $x^3$ as $x \to -\infty$:

   • As $x \to -\infty$, $x^3$ becomes very large in magnitude but negative, meaning $x^3 \to -\infty$.

2. Effect on $\frac{9}{x^3}$:

   • Since $x^3 \to -\infty$, the expression $\frac{9}{x^3}$ approaches $0$ but remains negative because the denominator is negative for negative $x$.

3. Conclusion:

   • The value of $\frac{9}{x^3}$ approaches 0 from the negative side as $x \to -\infty$.
   • Hence, the limit is 0.

$\lim_{x \to -\infty} \frac{9}{x^3} = 0$

This result tells us that as $x$ becomes very large in the negative direction, the function $\frac{9}{x^3}$ tends toward 0.

Would you like more details or further clarification on this? Here are some related questions you could explore:

1. What is the limit of $\frac{9}{x^2}$ as $x \to -\infty$?
2. How does the behavior of even powers of $x$ differ from odd powers as $x \to -\infty$?
3. What is the limit of $\frac{-9}{x^3}$ as $x \to -\infty$?
4. How would the limit change if $x \to \infty$ instead of $x \to -\infty$?
5. Can we use L'Hôpital's Rule to evaluate limits involving $\frac{9}{x^3}$?

Tip: As $x \to \infty$, rational functions with higher powers of $x$ in the denominator generally tend toward 0, regardless of the numerator.