Question

find the limit of $f(x)=\frac{-2 + \frac{9}{x}}{4-\frac{9}{x^{2}}}$ 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 $\frac{1}{x^n}$ as $x\to\pm\infty$

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

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

We have $f(x)=\frac{- 2+\frac{9}{x}}{4-\frac{9}{x^2}}$. Using the limit rules, $\lim_{x\to\infty}\frac{- 2+\frac{9}{x}}{4-\frac{9}{x^2}}=\frac{\lim_{x\to\infty}(-2)+\lim_{x\to\infty}\frac{9}{x}}{\lim_{x\to\infty}4-\lim_{x\to\infty}\frac{9}{x^2}}$. Since $\lim_{x\to\infty}\frac{9}{x}=0$ and $\lim_{x\to\infty}\frac{9}{x^2}=0$, we get $\frac{-2 + 0}{4-0}=-\frac{1}{2}$.

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

Similarly, for $x\to-\infty$, $\lim_{x\to-\infty}\frac{- 2+\frac{9}{x}}{4-\frac{9}{x^2}}=\frac{\lim_{x\to-\infty}(-2)+\lim_{x\to-\infty}\frac{9}{x}}{\lim_{x\to-\infty}4-\lim_{x\to-\infty}\frac{9}{x^2}}$. Since $\lim_{x\to-\infty}\frac{9}{x}=0$ and $\lim_{x\to-\infty}\frac{9}{x^2}=0$, we get $\frac{-2 + 0}{4-0}=-\frac{1}{2}$.

Answer:

$-\frac{1}{2}$