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 horizontal asymptote rules

For rational functions $f(x)=\frac{P(x)}{Q(x)}$ (where $P(x)$ is numerator, $Q(x)$ is denominator, both polynomials):

1. If $\deg(P) < \deg(Q)$: Horizontal asymptote $y=0$
2. If $\deg(P) = \deg(Q)$: Horizontal asymptote $y=\frac{\text{leading coefficient of } P}{\text{leading coefficient of } Q}$
3. If $\deg(P) > \deg(Q)$: No horizontal asymptote

Step2: Analyze Option 1

$f(x)=\frac{2x-1}{3x^2}$: $\deg(P)=1$, $\deg(Q)=2$. $\deg(P)<\deg(Q)$, so asymptote $y=0$.

Step3: Analyze Option 2

$f(x)=\frac{x-1}{3x}$: $\deg(P)=1$, $\deg(Q)=1$. $\deg(P)=\deg(Q)$, so asymptote $y=\frac{1}{3}$.

Step4: Analyze Option 3

$f(x)=\frac{2x^2}{3x-1}$: $\deg(P)=2$, $\deg(Q)=1$. $\deg(P)>\deg(Q)$, so no horizontal asymptote.

Step5: Analyze Option 4

$f(x)=\frac{3x^2}{x^2-1}$: $\deg(P)=2$, $\deg(Q)=2$. $\deg(P)=\deg(Q)$, so asymptote $y=\frac{3}{1}=3$.

Answer:

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