determine the limit shown below in simplest form. lim x→−1 -3x - 9 / x² - 1

Explanation:

Step1: Factor the expressions

Factor the numerator $-3x - 9=-3(x + 3)$ and the denominator $x^{2}-1=(x + 1)(x - 1)$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. So the limit becomes $\lim_{x\rightarrow - 1}\frac{-3(x + 3)}{(x + 1)(x - 1)}$.

Step2: Analyze the limit

As $x\rightarrow - 1$, the denominator $(x + 1)(x - 1)$ approaches $0$ and the numerator $-3(x + 3)$ approaches $-3(-1 + 3)=-6$. Since the numerator is non - zero and the denominator approaches $0$, the limit does not exist. We can also check the left - hand limit and right - hand limit. Left - hand limit: $\lim_{x\rightarrow - 1^{-}}\frac{-3(x + 3)}{(x + 1)(x - 1)}$. As $x\rightarrow - 1^{-}$, $(x + 1)\rightarrow0^{-}$ and $(x - 1)\rightarrow - 2$, and $-3(x + 3)\rightarrow - 6$. So $\lim_{x\rightarrow - 1^{-}}\frac{-3(x + 3)}{(x + 1)(x - 1)}=\infty$. Right - hand limit: $\lim_{x\rightarrow - 1^{+}}\frac{-3(x + 3)}{(x + 1)(x - 1)}$. As $x\rightarrow - 1^{+}$, $(x + 1)\rightarrow0^{+}$ and $(x - 1)\rightarrow - 2$, and $-3(x + 3)\rightarrow - 6$. So $\lim_{x\rightarrow - 1^{+}}\frac{-3(x + 3)}{(x + 1)(x - 1)}=-\infty$.

Answer:

The limit does not exist.