To solve the limit problem:

$\lim_{x \to 0} \frac{x + 3}{x^2(x - 4)}$

We will analyze the behavior of the expression as $x \to 0$.

Step 1: Analyze the numerator and denominator at $x = 0$

• The numerator $x + 3$ becomes $3$ as $x \to 0$.
• The denominator is $x^2(x - 4)$. At $x = 0$: $x^2(0 - 4) = 0^2 \cdot (-4) = 0$ So, the denominator approaches 0 as $x \to 0$.

Step 2: Behavior around 0

To understand the limit's behavior, we need to check the direction from both sides of 0:

1. As $x \to 0^+$ (from the right):

• Numerator: $x + 3 \to 3$ (positive).
• Denominator: $x^2$ is positive, and $(x - 4) \to -4$, so the denominator is negative (positive $x^2$ times negative $(x - 4)$). $\text{The fraction } \frac{3}{0^-} \text{ (positive numerator and negative denominator)} \to -\infty$

2. As $x \to 0^-$ (from the left):

• Numerator: $x + 3 \to 3$ (positive).
• Denominator: $x^2$ is still positive (because squaring a negative number gives a positive value), and $(x - 4) \to -4$, so the denominator is negative. $\text{The fraction } \frac{3}{0^-} \to -\infty$

Step 3: Conclusion

The limit approaches $-\infty$ from both sides as $x \to 0$.

Final Answer:

The limit is $-\infty$.

Do you have any questions or want further details?

Here are 5 questions to consider:

1. What is the definition of an infinite limit?
2. How does the sign of the denominator affect the limit as $x \to 0$?
3. Why is squaring a negative number always positive?
4. What is the behavior of functions involving $x^2$ as $x \to 0$?
5. How can limits help in analyzing discontinuities?

Tip: Always check the sign of both the numerator and denominator when evaluating limits approaching 0 to determine the direction of infinity.