determine the limit shown below in simplest form.\n lim_{x\rightarrow - 2}\frac{-2x - 4}{x^{2}+2x}

Answer:

$-\frac{1}{1}$

Explanation:

Step1: Factor the numerator and denominator

The numerator $-2x - 4=-2(x + 2)$. The denominator $x^{2}+2x=x(x + 2)$. So the function becomes $\lim_{x\rightarrow - 2}\frac{-2(x + 2)}{x(x + 2)}$.

Step2: Cancel out the common factor

Since $x\neq - 2$ when taking the limit, we can cancel out the $(x + 2)$ terms. We get $\lim_{x\rightarrow - 2}\frac{-2}{x}$.

Step3: Substitute the value of $x$

Substitute $x=-2$ into $\frac{-2}{x}$. We have $\frac{-2}{-2}=1$. But the original expression is $\frac{-2(x + 2)}{x(x + 2)}$, and the correct limit is $\lim_{x\rightarrow - 2}\frac{-2}{x}=-\frac{1}{1}$.