Question

find the limit of $f(x)=\frac{-7 + \frac{6}{x}}{9-\frac{6}{x^{2}}}$ as $x$ approaches $infty$ and as $x$ approaches $-infty$.
$lim_{x
ightarrowinfty}f(x)=square$ (type a simplified fraction.)

Explanation:

Step1: Recall limit rules for $\frac{1}{x}$

As $x\to\pm\infty$, $\lim_{x\to\pm\infty}\frac{1}{x}=0$ and $\lim_{x\to\pm\infty}\frac{1}{x^{2}} = 0$.

Step2: Find $\lim_{x\to\infty}f(x)$

We have $f(x)=\frac{-7+\frac{6}{x}}{9 - \frac{6}{x^{2}}}$. Using the limit rules for sums, differences and quotients of functions, $\lim_{x\to\infty}f(x)=\frac{\lim_{x\to\infty}(-7)+\lim_{x\to\infty}\frac{6}{x}}{\lim_{x\to\infty}9-\lim_{x\to\infty}\frac{6}{x^{2}}}$. Since $\lim_{x\to\infty}(-7)= - 7$, $\lim_{x\to\infty}\frac{6}{x}=0$, $\lim_{x\to\infty}9 = 9$ and $\lim_{x\to\infty}\frac{6}{x^{2}}=0$, then $\lim_{x\to\infty}f(x)=\frac{-7 + 0}{9-0}=-\frac{7}{9}$.

Step3: Find $\lim_{x\to-\infty}f(x)$

Similarly, $\lim_{x\to-\infty}f(x)=\frac{\lim_{x\to-\infty}(-7)+\lim_{x\to-\infty}\frac{6}{x}}{\lim_{x\to-\infty}9-\lim_{x\to-\infty}\frac{6}{x^{2}}}$. Since $\lim_{x\to-\infty}(-7)=-7$, $\lim_{x\to-\infty}\frac{6}{x}=0$, $\lim_{x\to-\infty}9 = 9$ and $\lim_{x\to-\infty}\frac{6}{x^{2}}=0$, then $\lim_{x\to-\infty}f(x)=\frac{-7+0}{9 - 0}=-\frac{7}{9}$.

Answer:

$-\frac{7}{9}$