Is $y=\frac{x^2}{x^3+3}$ has a horizontal asymptote?

Question

Is $y=\frac{x^2}{x^3+3}$ has a horizontal asymptote?

Since the graph passes through $(0,0)$, hence $y=0$ is not a horizontal asymptote. Can a rational function may not have a horizontal asymptote?

Answer 1

Hints:

An asymptote is a straight line to which the curve approaches while it moves away from the origin. The curve can approach the line from one side, or it can intersect it again and again. Not every curve which goes infinitely far from the origin (infinite branch of the curve) has an asymptote.

For functions given in explicit form $y=f(x)$, we know: the vertical asymptotes are at points of discontinuity where the function $f(x)$ has an infinite jump; the horizontal and oblique asymptotes have the equation:

$$y=kx+b, \space \text{ with } \space \space k=\lim_{x \to \infty}\frac{f(x)}{x}, \space \space b=\lim_{x \to \infty}\left[f(x)-kx \right].$$

Figure:

Answer 2

Yes, of course:

$$\lim_{x\rightarrow+\infty}\frac{x^2}{x^3+3}=\lim_{x\rightarrow+\infty}\frac{\frac{1}{x}}{1+\frac{3}{x^3}}=0,$$ $$\lim_{x\rightarrow-\infty}\frac{x^2}{x^3+3}=\lim_{x\rightarrow-\infty}\frac{\frac{1}{x}}{1+\frac{3}{x^3}}=0.$$ Id est, $y=0$ is a horizontal asymptote.

The function $f(x)=\frac{x^3}{x^2+3}$ has no a horizontal asymptote.