#### Answer

$-\dfrac{1}{2}$

#### Work Step by Step

We want to find $\lim\limits_{x \to -2}\dfrac{-2x-4}{x^3+2x^2}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -2 is zero, and if we try to substitute $x=-2$ directly, we get zero in both the numerator and the denominator. This means we have a common factor of $x+2$ in both the numerator and the denominator.
Note that $\dfrac{-2x-4}{x^3+2x^2}=\dfrac{-2(x+2)}{x^2(x+2)}=\dfrac{-2}{x^2}$ for all $x \neq -2.$
Thus $\lim\limits_{x \to -2}\dfrac{-2x-4}{x^3+2x^2}=\lim\limits_{x \to -2}\dfrac{-2}{x^2}=\dfrac{-2}{(-2)^2}=\dfrac{-2}{4}=-\dfrac{1}{2}.$