Evaluate the numerator as x 0^-.

As $x$ approaches 0 from the left, $x-1$ approaches $0-1$.

$$\lim_{x \to 0^-} (x-1) = -1$$

The denominator is $x^2(x+2)$.
As $x$ approaches 0 from the left, $x^2$ will be a very small positive number (e.g., $(-0.001)^2 = 0.000001$). So, $x^2 \to 0^+$.
As $x$ approaches 0 from the left, $x+2$ approaches $0+2$.

$$\lim_{x \to 0^-} x^2 = 0^+$$ $$\lim_{x \to 0^-} (x+2) = 2$$

Therefore, the denominator $x^2(x+2)$ approaches $(0^+)(2)$, which is $0^+$.

The limit becomes a negative number divided by a very small positive number.

$$\lim_{x \to 0^-} \frac{x-1}{x^2(x+2)} = \frac{-1}{0^+}$$

When a negative number is divided by a very small positive number, the result tends to negative infinity.

$$\frac{-1}{0^+} = -\infty$$

The final answer is $-\infty$.