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 a rational function \( f(x)=\frac{N(x)}{D(x)} \), where \( N(x) \) is the numerator and \( D(x) \) is the denominator (both polynomials):

• If the degree of \( N(x) \) (\( n \)) < degree of \( D(x) \) (\( d \)): Horizontal asymptote is \( y = 0 \).
• If \( n = d \): Horizontal asymptote is \( y=\frac{\text{leading coefficient of }N(x)}{\text{leading coefficient of }D(x)} \).
• If \( n > d \): No horizontal asymptote (may have an oblique/slant asymptote).

Step2: Analyze Each Option

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

Degree of numerator (\( n \)) = 1, Degree of denominator (\( d \)) = 2. \( n < d \), so horizontal asymptote \( y = 0 \).

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

Degree of numerator (\( n \)) = 1, Degree of denominator (\( d \)) = 1. \( n = d \), horizontal asymptote \( y=\frac{1}{3} \).

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

Degree of numerator (\( n \)) = 2, Degree of denominator (\( d \)) = 1. \( n > d \), so no horizontal asymptote.

Option 4: \( f(x)=\frac{3x^{2}}{x^{2}-1} \)

Degree of numerator (\( n \)) = 2, Degree of denominator (\( d \)) = 2. \( n = d \), horizontal asymptote \( y = 3 \).

Answer:

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