find the limit of the rational function (a) as $x\\to\\infty$ and (b) as $x\\to -\\infty$. write $\\infty$…
find the limit of the rational function (a) as $x\\to\\infty$ and (b) as $x\\to -\\infty$. write $\\infty$ or $-\\infty$ where appropriate.\n$f(x)=\\frac{4x^{3}+8}{x^{3}-x^{2}+x + 8}$\n a. $\\lim_{x\\to\\infty}(\\frac{4x^{3}+8}{x^{3}-x^{2}+x + 8})=\\square$\n(simplify your answer.)
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\rightarrow\infty}\frac{4x^{3}+8}{x^{3}-x^{2}+x + 8}&=\lim_{x\rightarrow\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\rightarrow\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 $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$, and $\lim_{x\rightarrow\infty}\frac{8}{x^{3}}=0$. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{4+\frac{8}{x^{3}}}{1-\frac{1}{x}+\frac{1}{x^{2}}+\frac{8}{x^{3}}}&=\frac{\lim_{x\rightarrow\infty}(4)+\lim_{x\rightarrow\infty}\frac{8}{x^{3}}}{\lim_{x\rightarrow\infty}(1)-\lim_{x\rightarrow\infty}\frac{1}{x}+\lim_{x\rightarrow\infty}\frac{1}{x^{2}}+\lim_{x\rightarrow\infty}\frac{8}{x^{3}}}\ &=\frac{4 + 0}{1-0 + 0+0}\ &=4 \end{align*} ]
Answer:
4