find the limit of the rational function (a) as x→∞ and (b) as x→ - ∞. write ∞ or - ∞ where appropriate. f(x)=\frac{5x^{4}+9}{x^{4}-x^{3}+x + 5} a. lim_{x
ightarrowinfty}(\frac{5x^{4}+9}{x^{4}-x^{3}+x + 5})=square (simplify your answer.)

Question

Accepted Answer

# 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.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{5x^{4}+9}{x^{4}-x^{3}+x + 5}&=\lim_{x
ightarrow\infty}\frac{\frac{5x^{4}}{x^{4}}+\frac{9}{x^{4}}}{\frac{x^{4}}{x^{4}}-\frac{x^{3}}{x^{4}}+\frac{x}{x^{4}}+\frac{5}{x^{4}}}\
&=\lim_{x
ightarrow\infty}\frac{5+\frac{9}{x^{4}}}{1-\frac{1}{x}+\frac{1}{x^{3}}+\frac{5}{x^{4}}}
\end{align*}
$$
## Step2: Use limit rules
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{5+\frac{9}{x^{4}}}{1-\frac{1}{x}+\frac{1}{x^{3}}+\frac{5}{x^{4}}}&=\frac{\lim_{x
ightarrow\infty}(5)+\lim_{x
ightarrow\infty}\frac{9}{x^{4}}}{\lim_{x
ightarrow\infty}(1)-\lim_{x
ightarrow\infty}\frac{1}{x}+\lim_{x
ightarrow\infty}\frac{1}{x^{3}}+\lim_{x
ightarrow\infty}\frac{5}{x^{4}}}\
&=\frac{5 + 0}{1-0 + 0+0}\
&=5
\end{align*}
$$
# Answer:
$5$

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QUESTION IMAGE

Question

Explanation:

Step1: Divide by highest - power of x

Step2: Use limit rules

Answer: