find the limit of the rational function (a) as x→∞ and (b) as x→ - ∞. write ∞ or - ∞ where appropriate…

Explanation:

Step1: Divide by highest - power of x

Divide both the numerator and denominator by $x^{4}$, the highest - power of $x$ in the denominator. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{5x^{4}+9}{x^{4}-x^{3}+x + 5}&=\lim_{x\rightarrow\infty}\frac{\frac{5x^{4}}{x^{4}}+\frac{9}{x^{4}}}{\frac{x^{4}}{x^{4}}-\frac{x^{3}}{x^{4}}+\frac{x}{x^{4}}+\frac{5}{x^{4}}}\ &=\lim_{x\rightarrow\infty}\frac{5+\frac{9}{x^{4}}}{1-\frac{1}{x}+\frac{1}{x^{3}}+\frac{5}{x^{4}}} \end{align*} ]

Step2: Use limit rules

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{5+\frac{9}{x^{4}}}{1-\frac{1}{x}+\frac{1}{x^{3}}+\frac{5}{x^{4}}}&=\frac{\lim_{x\rightarrow\infty}(5)+\lim_{x\rightarrow\infty}\frac{9}{x^{4}}}{\lim_{x\rightarrow\infty}(1)-\lim_{x\rightarrow\infty}\frac{1}{x}+\lim_{x\rightarrow\infty}\frac{1}{x^{3}}+\lim_{x\rightarrow\infty}\frac{5}{x^{4}}}\ &=\frac{5 + 0}{1-0 + 0+0}\ &=5 \end{align*} ]

Answer:

$5$