find $lim_{x\rightarrow - 2}\frac{x^{2}-2x - 8}{3x + 6}$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-2x - 8=(x + 2)(x - 4)$. The denominator $3x+6 = 3(x + 2)$. So the function becomes $\lim_{x\rightarrow - 2}\frac{(x + 2)(x - 4)}{3(x + 2)}$.

Step2: Cancel out the common factor

Cancel out the common factor $(x + 2)$ (since $x\neq - 2$ when taking the limit), we get $\lim_{x\rightarrow - 2}\frac{x - 4}{3}$.

Step3: Substitute the value of $x$

Substitute $x=-2$ into $\frac{x - 4}{3}$, we have $\frac{-2-4}{3}=\frac{-6}{3}=-2$.

Answer:

$-2$