Question

determine the horizontal asymptote of the function. if none exists, state that fact.

f(x)=\frac{3 x^{3}-3 x + 4}{15 x^{3}+3 x - 7}

select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

a. the function has one horizontal asymptote, (type an equation )
b. the function has two horizontal asymptotes. the top asymptote is and the bottom asymptote is (type equations )
c. the function has no horizontal asymptotes.

Explanation:

Step1: Divide numerator and denominator by \(x^3\)

$$\lim_{x ightarrow\pm\infty}\frac{3x^{3}-3x + 4}{15x^{3}+3x - 7}=\lim_{x ightarrow\pm\infty}\frac{3-\frac{3}{x^{2}}+\frac{4}{x^{3}}}{15+\frac{3}{x^{2}}-\frac{7}{x^{3}}}$$

Step2: Apply limit rules

As \(x
ightarrow\pm\infty\), \(\frac{1}{x^{n}}
ightarrow0\) for \(n>0\). So, \(\lim_{x
ightarrow\pm\infty}\frac{3-\frac{3}{x^{2}}+\frac{4}{x^{3}}}{15+\frac{3}{x^{2}}-\frac{7}{x^{3}}}=\frac{3 - 0+0}{15+0 - 0}\)

Answer:

A. The function has one horizontal asymptote, \(y = \frac{1}{5}\)