Question

determine the horizontal asymptote of the function. if none exists, state that fact.
f(x)=\frac{4 x}{x^{3}-3 x^{2}}
select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice
a. the function has two horizontal asymptotes. the top asymptote is ( square ) and the bottom asymptote is
(type equations.)
b. the function has one horizontal asymptote, ( square ). (type an equation.)
c. the function has no horizontal asymptotes

Explanation:

Step1: Analyze the degrees of numerator and denominator

The degree of the numerator \(n\) of \(y = \frac{4x}{x^{3}-3x^{2}}\) is \(n = 1\) (since the highest - power of \(x\) in the numerator is \(x^1\)), and the degree of the denominator \(m\) is \(m=3\) (since the highest - power of \(x\) in the denominator is \(x^3\)).

Step2: Apply the horizontal asymptote rule

The rule for horizontal asymptotes of a rational function \(y=\frac{f(x)}{g(x)}\) where \(f(x)=a_nx^n+\cdots\) and \(g(x)=b_mx^m+\cdots\) is:
If \(n\lt m\), then \(y = 0\) is the horizontal asymptote.
Since \(n = 1\) and \(m = 3\) (\(1\lt3\)), we find the limit as \(x\to\pm\infty\).
We use the fact that \(\lim_{x\to\pm\infty}\frac{4x}{x^{3}-3x^{2}}=\lim_{x\to\pm\infty}\frac{\frac{4x}{x^{3}}}{\frac{x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}}\) (divide numerator and denominator by \(x^{3}\)).

$$
LATEXBLOCK0
$$

As \(x\to\pm\infty\), \(\lim_{x\to\pm\infty}\frac{4}{x^{2}}=0\) and \(\lim_{x\to\pm\infty}\frac{3}{x}=0\). So \(\lim_{x\to\pm\infty}\frac{\frac{4}{x^{2}}}{1-\frac{3}{x}} = 0\)

Answer:

B. The function has one horizontal asymptote, \(y = 0\)