To find the limit:

$\lim_{x \to 0^-} \left( x^3 - \frac{1}{x^2} \right)$

we need to evaluate the behavior of the expression as $x$ approaches 0 from the left (negative side).

Breaking down the terms:

1. $x^3$ tends to 0 as $x \to 0$ because the cube of a small number approaches zero.
2. $\frac{1}{x^2}$ becomes very large as $x$ approaches 0 from the left. Since $x^2$ is always positive (whether $x$ is positive or negative), $\frac{1}{x^2}$ becomes infinitely large as $x \to 0^-$.

Combining the terms:

• The term $x^3$ approaches 0.
• The term $\frac{1}{x^2}$ approaches $+\infty$.

So, the limit is dominated by the $\frac{1}{x^2}$ term, which tends to $+\infty$.

Thus, the limit is:

$\lim_{x \to 0^-} \left( x^3 - \frac{1}{x^2} \right) = -\infty$

Summary:

As $x \to 0^-$, the expression tends to negative infinity because the negative reciprocal term dominates.

Would you like further clarification or more examples of limits?

Here are some related questions for you to explore:

1. What happens when you evaluate the limit $\lim_{x \to 0^+} \left( x^3 - \frac{1}{x^2} \right)$?
2. How would the limit change if we added a constant to the expression, such as $x^3 - \frac{1}{x^2} + 5$?
3. What is the behavior of the expression as $x \to \infty$?
4. How does the result change if we replace $x^3$ with $x^2$ in the original expression?
5. What is the limit of the function $\frac{1}{x^2}$ as $x \to 0^-$?

Tip: When dealing with limits involving powers and fractions, always pay attention to the dominant term as $x$ approaches the point of interest (in this case, $x = 0$).