#### Answer

$$\lim_{x\to0}\frac{-1}{x^2(x+1)}=-\infty$$

#### Work Step by Step

$$A=\lim_{x\to0}\frac{-1}{x^2(x+1)}=\frac{-1}{\lim_{x\to0}(x^2)(0+1)}=\frac{-1}{\lim_{x\to0}(x^2)}$$
As $x\to0$, $x$ approaches $0$ and $x^2$ approaches $0$ as well.
Since $x^2\ge0$ for all $x$, $-1/x^2\lt0$, so $-1/x^2$ will approach $-\infty$. Therefore,
$$A=-\infty$$