We are tasked with evaluating the limit:

$\lim_{x \to 0} \frac{x^3 - 1}{x^2}$

Let's solve this step by step using algebraic methods and limit theorems.

Step 1: Direct Substitution

First, attempt direct substitution by plugging $x = 0$ into the expression.

$\frac{0^3 - 1}{0^2} = \frac{-1}{0}$

Since division by zero is undefined, we cannot directly substitute $x = 0$. This suggests that we need a different approach, such as simplifying the expression.

Step 2: Rewrite the Expression

We can simplify the expression by separating the terms of the fraction:

$\frac{x^3 - 1}{x^2} = \frac{x^3}{x^2} - \frac{1}{x^2}$

This simplifies further to:

$x - \frac{1}{x^2}$

Step 3: Analyze the Behavior as $x \to 0$

Now, evaluate the limit for each term separately.

• The first term, $x$, clearly approaches 0 as $x \to 0$.
• The second term, $-\frac{1}{x^2}$, approaches $-\infty$ as $x \to 0$ because $\frac{1}{x^2}$ grows without bound when $x$ is very small and close to 0.

Thus, the entire expression:

$x - \frac{1}{x^2}$

will approach $-\infty$ as $x \to 0$.

Step 4: Conclusion

Therefore, the limit does not exist in the usual sense because it tends to $-\infty$, which indicates that the function diverges as $x$ approaches 0.

$\boxed{\lim_{x \to 0} \frac{x^3 - 1}{x^2} = -\infty}$

Would you like to go into more details, or have any questions?

Here are 5 related questions for further exploration:

1. What is the limit of $\frac{x^3 - 1}{x^2}$ as $x$ approaches infinity?
2. How do limits involving infinity differ from finite limits?
3. What is the behavior of $\frac{x^n - 1}{x^m}$ for general $n$ and $m$ as $x$ approaches zero?
4. How does one handle indeterminate forms like $\frac{0}{0}$ using L'Hopital's Rule?
5. How can we graphically interpret the behavior of this function near zero?

Tip: When encountering limits that tend toward infinity or indeterminate forms, simplifying the function or using tools like L'Hopital's Rule can often clarify the behavior of the function.