find the limit of f(x) = \\frac{9}{x}-12 as x approaches ∞ and as x approaches -∞.\n\\lim_{x\\to\\infty}f(x)=…

find the limit of f(x) = \\frac{9}{x}-12 as x approaches ∞ and as x approaches -∞.\n\\lim_{x\\to\\infty}f(x)=\\square\n(type a simplified fraction.)
Answer
Explanation:
Step1: Recall limit rule for $\frac{c}{x}$
We know that $\lim_{x\rightarrow\pm\infty}\frac{c}{x}=0$ for a constant $c$. Here $c = 9$.
Step2: Find $\lim_{x\rightarrow\infty}f(x)$
$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}(\frac{9}{x}-12)$. Using the limit - difference rule $\lim_{x\rightarrow a}(u(x)-v(x))=\lim_{x\rightarrow a}u(x)-\lim_{x\rightarrow a}v(x)$, we have $\lim_{x\rightarrow\infty}\frac{9}{x}-\lim_{x\rightarrow\infty}12$. Since $\lim_{x\rightarrow\infty}\frac{9}{x}=0$ and $\lim_{x\rightarrow\infty}12 = 12$, then $\lim_{x\rightarrow\infty}f(x)=0 - 12=-12$.
Step3: Find $\lim_{x\rightarrow-\infty}f(x)$
$\lim_{x\rightarrow-\infty}f(x)=\lim_{x\rightarrow-\infty}(\frac{9}{x}-12)$. Using the limit - difference rule $\lim_{x\rightarrow a}(u(x)-v(x))=\lim_{x\rightarrow a}u(x)-\lim_{x\rightarrow a}v(x)$, we have $\lim_{x\rightarrow-\infty}\frac{9}{x}-\lim_{x\rightarrow-\infty}12$. Since $\lim_{x\rightarrow-\infty}\frac{9}{x}=0$ and $\lim_{x\rightarrow-\infty}12 = 12$, then $\lim_{x\rightarrow-\infty}f(x)=0 - 12=-12$.
Answer:
-12