To determine which function has no horizontal asymptote, we need to analyze the degrees of the numerator and the denominator in each function and apply the rules for finding horizontal asymptotes.
1. \( f(x) = \frac{2x - 1}{3x^2} \)
- The degree of the numerator: \(1\) (since the highest power of \(x\) is \(x^1\))
- The degree of the denominator: \(2\) (since the highest power of \(x\) is \(x^2\))
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).
2. \( f(x) = \frac{x - 1}{3x} \)
- The degree of the numerator: \(1\) (since the highest power of \(x\) is \(x^1\))
- The degree of the denominator: \(1\) (since the highest power of \(x\) is \(x^1\))
- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is \( y = \) (leading coefficient of the numerator) / (leading coefficient of the denominator). Therefore, the horizontal asymptote here is \( y = \frac{1}{3} \).
3. \( f(x) = \frac{2x^2}{3x - 1} \)
- The degree of the numerator: \(2\) (since the highest power of \(x\) is \(x^2\))
- The degree of the denominator: \(1\) (since the highest power of \(x\) is \(x^1\))
- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
4. \( f(x) = \frac{3x^2}{x^2 - 1} \)
- The degree of the numerator: \(2\) (since the highest power of \(x\) is \(x^2\))
- The degree of the denominator: \(2\) (since the highest power of \(x\) is \(x^2\))
- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is \( y = \) (leading coefficient of the numerator) / (leading coefficient of the denominator). Therefore, the horizontal asymptote here is \( y = \frac{3}{1} = 3 \).
From this analysis, we see that the function \( f(x) = \frac{2x^2}{3x - 1} \) has no horizontal asymptote.
Therefore, the function that has no horizontal asymptote is:
[tex]\[ \boxed{f(x) = \frac{2x^2}{3x - 1}} \][/tex]
