To determine which function has no horizontal asymptote, we need to look at the behavior of each function as [tex]\( x \)[/tex] approaches infinity ([tex]\( x \to \infty \)[/tex]).
### 1. [tex]\( f(x) = \frac{2x - 1}{3x^2} \)[/tex]
To find the horizontal asymptote, we look at the limits as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{2x - 1}{3x^2} \][/tex]
Dividing the numerator and the denominator by [tex]\( x^2 \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{\frac{2}{x} - \frac{1}{x^2}}{3} = \frac{0}{3} = 0 \][/tex]
So, [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
### 2. [tex]\( f(x) = \frac{x - 1}{3x} \)[/tex]
To find the horizontal asymptote:
[tex]\[ \lim_{x \to \infty} \frac{x - 1}{3x} \][/tex]
Dividing the numerator and the denominator by [tex]\( x \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{1 - \frac{1}{x}}{3} = \frac{1}{3} \][/tex]
So, [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = \frac{1}{3} \)[/tex].
### 3. [tex]\( f(x) = \frac{2x^2}{3x - 1} \)[/tex]
To find the horizontal asymptote:
[tex]\[ \lim_{x \to \infty} \frac{2x^2}{3x - 1} \][/tex]
Dividing the numerator and the denominator by [tex]\( x \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{2x^2}{3x} = \lim_{x \to \infty} \frac{2x}{3} = \infty \][/tex]
Since the limit is infinity, [tex]\( f(x) \)[/tex] does not have a horizontal asymptote.
### 4. [tex]\( f(x) = \frac{3x^2}{x^2 - 1} \)[/tex]
To find the horizontal asymptote:
[tex]\[ \lim_{x \to \infty} \frac{3x^2}{x^2 - 1} \][/tex]
Dividing the numerator and the denominator by [tex]\( x^2 \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{3}{1 - \frac{1}{x^2}} = \frac{3}{1 - 0} = 3 \][/tex]
So, [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y = 3 \)[/tex].
### Conclusion
The function that has no horizontal asymptote is:
[tex]\[ f(x) = \frac{2x^2}{3x - 1} \][/tex]
