Chapter 1 - Functions and Limits - 1.5 The Limit of a Function - 1.5 Exercises - Page 61: 34

$-\infty$

$x\rightarrow 0$ means that the values of x approach 0 from either side. For any of these x, the numerator is negative (approaches $-1).$ The denominator is positive, because $x^{2}$is not negative and approaches 0, $(x+2)$ is positive and approaches +2, so their product is positive and approaches 0. So $\displaystyle \lim_{x\rightarrow 0}\frac{x-1}{x^{2}(x+2)}=-\infty$