Question

find the limit of f(x) = \frac{9}{x}-12 as x approaches ∞ and as x approaches -∞.
\lim_{x\to\infty}f(x)=\square
(type a simplified fraction.)

Explanation:

Step1: Recall limit rule for $\frac{c}{x}$

We know that $\lim_{x
ightarrow\pm\infty}\frac{c}{x}=0$ for a constant $c$. Here $c = 9$.

Step2: Find $\lim_{x

ightarrow\infty}f(x)$
$\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}(\frac{9}{x}-12)$. Using the limit - difference rule $\lim_{x
ightarrow a}(u(x)-v(x))=\lim_{x
ightarrow a}u(x)-\lim_{x
ightarrow a}v(x)$, we have $\lim_{x
ightarrow\infty}\frac{9}{x}-\lim_{x
ightarrow\infty}12$. Since $\lim_{x
ightarrow\infty}\frac{9}{x}=0$ and $\lim_{x
ightarrow\infty}12 = 12$, then $\lim_{x
ightarrow\infty}f(x)=0 - 12=-12$.

Step3: Find $\lim_{x

ightarrow-\infty}f(x)$
$\lim_{x
ightarrow-\infty}f(x)=\lim_{x
ightarrow-\infty}(\frac{9}{x}-12)$. Using the limit - difference rule $\lim_{x
ightarrow a}(u(x)-v(x))=\lim_{x
ightarrow a}u(x)-\lim_{x
ightarrow a}v(x)$, we have $\lim_{x
ightarrow-\infty}\frac{9}{x}-\lim_{x
ightarrow-\infty}12$. Since $\lim_{x
ightarrow-\infty}\frac{9}{x}=0$ and $\lim_{x
ightarrow-\infty}12 = 12$, then $\lim_{x
ightarrow-\infty}f(x)=0 - 12=-12$.

Answer:

-12