which function has no horizontal asymptote?\n f(x)=\frac{2x - 1}{3x^{2}}\n f(x)=\frac{x - 1}{3x}\n…

Explanation:

Step1: Recall horizontal - asymptote rules

For a rational function $f(x)=\frac{a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_0}{b_mx^m + b_{m - 1}x^{m - 1}+\cdots+b_0}$, if $n\lt m$, the horizontal asymptote is $y = 0$; if $n=m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$; if $n\gt m$, there is no horizontal asymptote.

Step2: Analyze the first function

For $f(x)=\frac{2x - 1}{3x^2}$, $n = 1$ and $m = 2$. Since $n\lt m$, the horizontal asymptote is $y = 0$.

Step3: Analyze the second function

For $f(x)=\frac{x - 1}{3x}$, $n = 1$ and $m = 1$. Then $y=\frac{1}{3}$ is the horizontal asymptote.

Step4: Analyze the third function

For $f(x)=\frac{2x^2}{3x - 1}$, $n = 2$ and $m = 1$. Since $n\gt m$, there is no horizontal asymptote.

Step5: Analyze the fourth function

For $f(x)=\frac{3x^2}{x^2 - 1}$, $n = 2$ and $m = 2$. Then $y = 3$ is the horizontal asymptote.

Answer:

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