Let's evaluate the function $f(x) = \frac{9}{x^3 - 1}$ as $x$ approaches 1 from the left (denoted as $x \to 1^-$) and from the right (denoted as $x \to 1^+$).

1. Understanding the function:

The function is given by:

$f(x) = \frac{9}{x^3 - 1}$

The expression $x^3 - 1$ can be factored using the difference of cubes:

$x^3 - 1 = (x - 1)(x^2 + x + 1)$

Thus, the function can be rewritten as:

$f(x) = \frac{9}{(x - 1)(x^2 + x + 1)}$

2. Evaluating the limit as $x \to 1^-$:

As $x$ approaches 1 from the left (values slightly less than 1), the factor $(x - 1)$ is slightly negative. The factor $(x^2 + x + 1)$ is always positive for real $x$.

So, as $x$ approaches 1 from the left:

• $(x - 1)$ approaches 0 from the negative side.
• $(x^2 + x + 1)$ approaches 3 (since $1^2 + 1 + 1 = 3$).

This makes the function:

$f(x) \approx \frac{9}{\text{(a small negative number)} \times 3} \approx -\infty$

So,

$\lim_{x \to 1^-} f(x) = -\infty$

3. Evaluating the limit as $x \to 1^+$:

As $x$ approaches 1 from the right (values slightly greater than 1), the factor $(x - 1)$ is slightly positive. The factor $(x^2 + x + 1)$ is again always positive.

So, as $x$ approaches 1 from the right:

• $(x - 1)$ approaches 0 from the positive side.
• $(x^2 + x + 1)$ approaches 3.

This makes the function:

$f(x) \approx \frac{9}{\text{(a small positive number)} \times 3} \approx +\infty$

So,

$\lim_{x \to 1^+} f(x) = +\infty$

Final Answer:

• $\lim_{x \to 1^-} f(x) = -\infty$
• $\lim_{x \to 1^+} f(x) = +\infty$

These results indicate that the function has a vertical asymptote at $x = 1$, where the function behavior is sharply different on either side of the asymptote.

Would you like to explore these limits further, or do you have any additional questions?

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Here are 5 related questions for further exploration:

1. How would the function behave as $x$ approaches 1 from values far from 1?
2. Can you find the limit of $f(x)$ as $x$ approaches 0 or infinity?
3. What happens to the function if you change the exponent in the denominator from 3 to 2?
4. How would the limit change if the function were $\frac{9}{x^3 + 1}$ instead?
5. Can you plot the graph of this function to visually confirm the asymptotic behavior?

Tip: When dealing with limits involving asymptotes, always consider the sign and magnitude of the components in the function as they approach critical values.