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 $f(x)=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}$, if $n\lt m$, the horizontal asymptote is $y = 0$; if $n=m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$; if $n>m$, 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 $n\lt m$, the horizontal asymptote is $y = 0$.

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

Here, $n = 1$ and $m = 1$. Then $y=\frac{1}{3}$ is the horizontal asymptote.

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

Here, $n = 2$ and $m = 1$. 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 $y = 3$ is the horizontal asymptote.

Answer:

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