Question

which function has no horizontal asymptote?

$f(x)=\frac{2x - 1}{3x^{2}}$

$f(x)=\frac{x - 1}{3x}$

$f(x)=\frac{2x^{2}}{3x - 1}$

$f(x)=\frac{3x^{2}}{x^{2}-1}$

Explanation:

Step1: Recall the rules for horizontal asymptotes

For a rational function \(f(x)=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\), if \(n < m\), \(y = 0\) is the horizontal asymptote; if \(n=m\), \(y=\frac{a_n}{b_m}\) is the horizontal asymptote; if \(n>m\), there is no horizontal asymptote.

Step2: Analyze \(f(x)=\frac{2x - 1}{3x^2}\)

Here \(n = 1\) (degree of numerator) and \(m=2\) (degree of denominator). Since \(n

Step3: Analyze \(f(x)=\frac{x - 1}{3x}\)

Here \(n = 1\) and \(m = 1\). Then \(y=\frac{1}{3}\) is the horizontal asymptote (because \(a_n = 1\), \(b_m=3\)).

Step4: Analyze \(f(x)=\frac{2x^2}{3x - 1}\)

Here \(n = 2\) (degree of numerator) and \(m = 1\) (degree of denominator). Since \(n>m\), there is no horizontal asymptote.

Step5: Analyze \(f(x)=\frac{3x^2}{x^2-1}\)

Here \(n = 2\) and \(m = 2\). Then \(y=\frac{3}{1}=3\) is the horizontal asymptote (because \(a_n = 3\), \(b_m = 1\)).

Answer:

\(f(x)=\frac{2x^2}{3x - 1}\)