Answer
$\lim\limits_{x \to \ -2}\frac{x^2-4}{x^3+2x^2}=\lim\limits_{x \to \ -2}\frac{x-2}{x^2}=\frac{-2-2}{(-2)^2}=\frac{-4}{4}=-1$
Work Step by Step
If we want to calculate the value of $\lim\limits_{x \to \ -2}\frac{x^2-4}{x^3+2x^2}$,
Here, we cannot use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, because it would mean that:
$\lim\limits_{x \to \ -2}\frac{x^2-4}{x^3+2x^2}=\frac{(-2)^2-4}{(-2)^3+2(-2)^2=\frac{0}{0}}$ which cannot be a limit.
Therefore, we have to simplify the fraction:
$\frac{x^2-4}{x^3+2x^2}=\frac{(x+2)(x-2)}{(x+2)(x^2)}=\frac{x-2}{x^2}$
If we substitute this function into the limit, we get:
$\lim\limits_{x \to \ -2}\frac{x^2-4}{x^3+2x^2}=\lim\limits_{x \to \ -2}\frac{x-2}{x^2}=\frac{-2-2}{(-2)^2}=\frac{-4}{4}=-1$
