Question

find the limit of the rational function (a) as $x\to\infty$ and (b) as $x\to -\infty$. write $\infty$ or $-\infty$ where appropriate.
$f(x)=\frac{4x^{3}+8}{x^{3}-x^{2}+x + 8}$
a. $\lim_{x\to\infty}(\frac{4x^{3}+8}{x^{3}-x^{2}+x + 8})=\square$
(simplify your answer.)

Explanation:

Step1: Divide by highest - power of x

Divide both the numerator and denominator by $x^{3}$, since the highest - power of $x$ in the denominator is $x^{3}$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{4x^{3}+8}{x^{3}-x^{2}+x + 8}&=\lim_{x ightarrow\infty}\frac{\frac{4x^{3}}{x^{3}}+\frac{8}{x^{3}}}{\frac{x^{3}}{x^{3}}-\frac{x^{2}}{x^{3}}+\frac{x}{x^{3}}+\frac{8}{x^{3}}}\\ &=\lim_{x ightarrow\infty}\frac{4+\frac{8}{x^{3}}}{1-\frac{1}{x}+\frac{1}{x^{2}}+\frac{8}{x^{3}}} \end{align*}$$

\]

Step2: Apply limit rules

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0$, and $\lim_{x
ightarrow\infty}\frac{8}{x^{3}}=0$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{4+\frac{8}{x^{3}}}{1-\frac{1}{x}+\frac{1}{x^{2}}+\frac{8}{x^{3}}}&=\frac{\lim_{x ightarrow\infty}(4)+\lim_{x ightarrow\infty}\frac{8}{x^{3}}}{\lim_{x ightarrow\infty}(1)-\lim_{x ightarrow\infty}\frac{1}{x}+\lim_{x ightarrow\infty}\frac{1}{x^{2}}+\lim_{x ightarrow\infty}\frac{8}{x^{3}}}\\ &=\frac{4 + 0}{1-0 + 0+0}\\ &=4 \end{align*}$$

\]

Answer:

4